What are the values and types of the critical points, if any, of #f(x,y)=y^2-x^2+3x-sqrt(2xy-4x)#?

Question

Cameron Alexander · Accepted Answer

A saddle point at

#x_0 = 1.43882, y_0 = 2.04309# Given #f(x,y)=y^2-x^2+3x-sqrt(2xy-4x)# The stationary points are found solving #grad f(x,y) = vec 0# or #{ (3 - 2 x - (2 y-4)/(2 sqrt[2 x y-4x])=0), (2 y - x/sqrt[2 x y-4x]=0) :}# Whose real solution(s) is #x_0 = 1.43882, y_0 = 2.04309# The qualification is made calculating the Hessian matrix in this point. #H(x,y) = grad(grad f(x,y)) = ((sqrt[x (y-2)]/(2 sqrt[2] x^2)-2, -1/( 2 sqrt[2] sqrt[x (y-2)])),(-1/(2 sqrt[2] sqrt[x (y-2)]), 2 + x^2/(2 x y-4x)^(3/2)))# so #H(x_0,y_0) = ((-1.95748, -1.41998),(-1.41998, 49.4181))# with characteristic polynomial #p(lambda) = lambda^2-"trace"(H)lambda+det(H)# with roots #lambda = -1.99669,lambda =49.4573# The roots are non null with oposite signs so the stationary point found, is a saddle point.

Christopher Allers · Answer

undefined To find the critical points of $ f(x,y) = y^2 - x^2 + 3x - \sqrt{2xy - 4x} $, we need to first find the partial derivatives with respect to $ x $ and $ y $ and then solve for $ f_x = 0 $ and $ f_y = 0 $.

$$
f_x = -2x + 3 - \frac{2y - 4}{2\sqrt{2xy - 4x}} = 0
$$
$$
f_y = 2y - \frac{2x}{\sqrt{2xy - 4x}} = 0
$$

Solving these equations simultaneously, we can find the critical points. However, the algebraic manipulation for these equations is complex and not suitable for a simple text response. You would need to solve the system of equations $ f_x = 0 $ and $ f_y = 0 $ to find the critical points.

Once you have the critical points, you can determine their values and types (whether they are maximum, minimum, or saddle points) by using the second partial derivative test or other appropriate methods.