Question

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

Answer 1

A saddle point at

`x_0 = 1.43882, y_0 = 2.04309`

Explanation:

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.