Question

Find the local maximum and minimum values and saddle point(s) of the function.

Answer 1

$f(x,y)=9e^y(y^2-x^2)$
$\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle$

Critical points occur where the gradient is zero. This is guaranteed whenever $x=0$ and either $y=0$ or $y=-2$.

The Hessian matrix for this function looks like

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}$

and has determinant

$|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)$

Maxima occur whenever the determinant is positive and $f_{xx}<0$. Minima occur whenever both the determinant and $f_{xx}$ are positive. Saddle points occur whenever the determinant is negative.

At $(0,0)$, you have a saddle point since the determinant reduces to -324, so $(0,0,0)$ is the saddle point.

At $(0,-2)$, the determinant is $\dfrac{324}{e^4}>0$ and $f_{xx}(0,-2)=-\dfrac{18}{e^2}<0$, so $\left(0,-2,\dfrac{36}{e^2}\right)$ is a local maximum.

No other critical points remain, so you're done.