How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for #f(x, y) = x^2 + 4x + y^2#?

Question

Emilia Aguilar · Accepted Answer

The unique critical point is #(x,y)=(-2,0)# and it is a local minimum (and the local minimum value of the function is #f(-2,0)#=4-8+0=-4#.

The first-order partial derivatives of #z=f(x,y)=x^2+4x+y^2# are #(partial z)/(partial x)=2x+4# and #(partial z)/(partial y)=2y#. Setting both of these equal to zero results in a system of equations whose unique solution is clearly #(x,y)=(-2,0)#, so this is the unique critical point of #f#. The second-order partials are #(partial^{2}z)/(partial x^{2})=2#, #(partial^{2}z)/(partial y^{2})=2#, and #(partial^{2}z)/(partial x partial y)=(partial^{2}z)/(partial y partial x)=0# As a result, the Second Derivative Test's discriminant (multivariable) equals #D=(partial^{2}z)/(partial x^{2})*(partial^{2}z)/(partial y^{2})-((partial^{2}z)/(partial x partial y))^2=2*2-0^2=4>0#, which indicates that the critical point is not a saddle point but rather either a local max or a local min. Since #(partial^{2}z)/(partial x^{2})=2>0#, the critical point is a local min.

Charles Anctil · Answer

See Explanation

#f(x, y) = x^2 + 4x + y^2# #f_x(x,y) = 2x+4# #f_y(x,y) = 2y# The critical point makes both partial derivatives #0# (simultaneously). For this function there is one critical point: #(-2,0)# To determine whether #f# has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.) #f_(x x)(x,y) = 2# #f_(x y)(x,y) = 0" "# (As usual, this is also #f_(y x)(x,y)#) #f_(y y)(x,y) = 2# At the critical point, evaluate the second partials (they are all constant in this case, but generally we can't skip this step). At the critical point #(-2,0)#, we get #A = f_(x x)(-2,0) = 2# #B = f_(x y)(-2,0)= 0# #C = f_(y y)(-2,0) = 2# Calculate #D = AC-B^2# #D = (2)(2)-(0)^2 = 4# Use the test for the second derivative: Since #D# is positive, we look at #A# and with #D > 0# and #A > 0#, we have a local minimum at the critical point. #f(-2,0) = 4 - 8 = -4# To sum up: #f# has a local minimum of #-4# at #(-2,0)#.

Christopher Allers · Answer

undefined To find the critical point, we first need to find the partial derivatives of the function $ f(x, y) $:

$ \frac{{\partial f}}{{\partial x}} = 2x + 4 $

$ \frac{{\partial f}}{{\partial y}} = 2y $

Setting both partial derivatives equal to zero to find critical points:

$ 2x + 4 = 0 \Rightarrow x = -2 $

$ 2y = 0 \Rightarrow y = 0 $

So, the critical point is $ (-2, 0) $.

To determine whether it's a local maximum, local minimum, or neither, we use the second partial derivative test.

$ \frac{{\partial^2 f}}{{\partial x^2}} = 2 $

$ \frac{{\partial^2 f}}{{\partial y^2}} = 2 $

$ \frac{{\partial^2 f}}{{\partial x \partial y}} = 0 $

At the critical point $ (-2, 0) $:

$ 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 = (2)(2) - (0)^2 = 4 $

Since $ D > 0 $ and $ \frac{{\partial^2 f}}{{\partial x^2}} > 0 $, the critical point $ (-2, 0) $ is a local minimum.