What are the extrema and saddle points of #f(x,y) = 2x^(2) + (xy)^2 + 5x^2 - y/x#?

Question

Ezra Adams · Accepted Answer

This function has no stationary points (are you sure that #f(x,y)=2x^2+(xy)^2+5x^2−y/x# is the one that you wanted to study?!).

According to the most diffused definition of saddle points (stationary points that are not extrema), you are searching for the stationary points of the function in its domain #D={(x,y) in RR^2 | x ne 0}=RR^2 setminus {(0,y) in RR^2}#. We can now rewrite the expression given for #f# in the following way: #f(x,y)=7x^2+x^2y^2-y/x# The way to identify them is to search for the points that nullify the gradient of #f#, which is the vector of the partial derivatives: #nabla f = ((del f)/(del x),(del f)/(del y))# Since the domain is an open set, we don't need to search for extrema eventually lying on the boundary, because an open sets contain no boundary points. So let's compute the gradient of the function: #nabla f (x,y)=(14x+2xy^2+y/x^2,2x^2y-1/x)# This is null when the following equations are simultaneously satisfied: #14x+2xy^2+y/x^2=0# #2x^2y=1/x# We can turn the second into #y=1/(2x^3)# and substitute it into the first to get #14x+2x(1/(2x^3))^2+(1/(2x^3))/x^2=0# #14x+1/(2x^5)+1/(2x^5)=0# #14x^6+1=0# This can't be satisfied for #x in RR#, so the gradient is never null on the domain. This means that the function has no stationary points!

Scarlett Allison · Answer

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Then, we determine the nature of these criticalTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical pointsTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the secondTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partialTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative testTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} =To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ andTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \fracTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{yTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \fracTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \fracTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial fTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} =To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} =To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xyTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 +To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdotTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xyTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. \[ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2.To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. SetTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set bothTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivativesTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equalTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal toTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zeroTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero andTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solveTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve forTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} =To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ xTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xyTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ yTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to findTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find criticalTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical pointsTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting theseTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3.To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal toTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. AfterTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero,To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After findingTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding theTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the criticalTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical pointsTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $.To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points,To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. AfterTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, computeTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After findingTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute theTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the secondTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points,To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-orderTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partialTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivativesTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative testTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives toTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test toTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classifyTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determineTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify themTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine ifTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maximaTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each criticalTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point isTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minimaTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is aTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima,To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximumTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, orTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum,To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddleTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimumTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle pointsTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, orTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle pointTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. TheTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The secondTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \fracTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partialTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivativesTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives ofTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 fTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y)To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ areTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} =To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 +To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2yTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 -To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \fracTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} =To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2yTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 +To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. \[ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \fracTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 fTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial yTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2} = To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2} = 4To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2} = 4xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2} = 4x^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ \[ \frac{\partial^2 f}{\partial y^2} = 4x^2 \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xyTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \fracTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial xTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partialTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial yTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y}To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y} =To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y} = To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y} = 4To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial xTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y} = 4xyTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ \[ \frac{\partial^2 f}{\partial x \partial y} = 4xy \To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partialTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial yTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} =To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4.To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. EvaluateTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate theTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xyTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminantTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ WeTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ DTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We thenTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D =To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluateTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate theseTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \fracTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partialTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivativesTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives atTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 fTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point andTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and useTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial xTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminantTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2}To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ DTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdotTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D =To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \fracTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xxTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}fTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yyTo find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 fTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy}To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 \To find the extrema and saddle points of \( f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial yTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ toTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determineTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2}To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the natureTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} -To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature ofTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the criticalTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical pointsTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial yTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. PleaseTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\rightTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please noteTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations toTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to findTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ atTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find theTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at eachTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the criticalTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each criticalTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical pointsTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical pointTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points andTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classifyTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. -To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them asTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - IfTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maximaTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ DTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima,To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D >To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minimaTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, orTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $To find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddleTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ andTo find the extrema and saddle points of $ f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle pointsTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points mayTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may beTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \fracTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthyTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthy andTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\partialTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthy and involveTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\partial^To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthy and involve solvingTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\partial^2To find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthy and involve solving equationsTo find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\partial^2 fTo find the extrema and saddle points of \( f(x,y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, we need to first find the critical points by taking the partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $. Then, we determine the nature of these critical points using the second partial derivative test. The partial derivatives of $ f(x,y) $ with respect to $ x $ and $ y $ are: $ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x^2 + \frac{y}{x^2} $ $ \frac{\partial f}{\partial y} = 2xy^3 - \frac{1}{x} $ Setting these partial derivatives equal to zero, we solve for $ x $ and $ y $. After finding the critical points, we use the second partial derivative test to determine if each critical point is a maximum, minimum, or saddle point. The second partial derivatives of $ f(x,y) $ are: $ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $ $ \frac{\partial^2 f}{\partial y^2} = 6xy $ $ \frac{\partial^2 f}{\partial x \partial y} = 4xy $ We then evaluate these second partial derivatives at each critical point and use the discriminant $ D = f_{xx}f_{yy} - (f_{xy})^2 $ to determine the nature of the critical points. Please note that the calculations to find the critical points and classify them as maxima, minima, or saddle points may be lengthy and involve solving equations.To find the extrema and saddle points of $ f(x, y) = 2x^2 + (xy)^2 + 5x^2 - \frac{y}{x} $, follow these steps: 1. Compute the first-order partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $. $$ \frac{\partial f}{\partial x} = 4x + 2xy^2 + 10x + \frac{y}{x^2} $$ $$ \frac{\partial f}{\partial y} = 2x^2 \cdot 2xy - \frac{1}{x} $$ 2. Set both partial derivatives equal to zero and solve for $ x $ and $ y $ to find critical points. 3. After finding the critical points, compute the second-order partial derivatives to classify them as maxima, minima, or saddle points. $$ \frac{\partial^2 f}{\partial x^2} = 4 + 2y^2 - \frac{2y}{x^3} $$ $$ \frac{\partial^2 f}{\partial y^2} = 4x^2 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4xy $$ 4. Evaluate the discriminant $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $ at each critical point. - If $ D > 0 $ and $ \frac{\partial^2 f}{\partial x^2} > 0 $, it's a local minimum. - If $ D > 0 $ and $ \frac{\partial^2 f}{\partial x^2} < 0 $, it's a local maximum. - If $ D < 0 $, it's a saddle point. - If $ D = 0 $, the test is inconclusive. 5. Substitute the critical points into the original function $ f(x, y) $ to determine the corresponding function values. These steps will provide you with information about the extrema and saddle points of the given function.