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

Sep 3, 2016

${x}_{0} = 1.43882 , {y}_{0} = 2.04309$

#### Explanation:

Given $f \left(x , y\right) = {y}^{2} - {x}^{2} + 3 x - \sqrt{2 x y - 4 x}$

The stationary points are found solving

$\nabla f \left(x , y\right) = \vec{0}$

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 x^2)-2, -1/( 2 sqrt sqrt[x (y-2)])),(-1/(2 sqrt sqrt[x (y-2)]), 2 + x^2/(2 x y-4x)^(3/2)))

so

$H \left({x}_{0} , {y}_{0}\right) = \left(\begin{matrix}- 1.95748 & - 1.41998 \\ - 1.41998 & 49.4181\end{matrix}\right)$

with characteristic polynomial

$p \left(\lambda\right) = {\lambda}^{2} - \text{trace} \left(H\right) \lambda + \det \left(H\right)$

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.