# What are the extrema and saddle points of f(x,y) = e^y(y^2-x^2)?

Jul 2, 2016

#### Answer:

$\left\{0 , 0\right\}$ saddle point
$\left\{0 , - 2\right\}$ local maximum

#### Explanation:

$f \left(x , y\right) = {e}^{y} \left({y}^{2} - {x}^{2}\right)$

so the sationary points are determined by solving

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

or

{ (-2 e^y x = 0), (2 e^y y + e^y (-x^2 + y^2) = 0) :}

giving two solutions

$\left(\begin{matrix}x = 0 & y = 0 \\ x = 0 & y = - 2\end{matrix}\right)$

Those points are qualified using

$H = \nabla \left(\nabla f \left(x , y\right)\right)$

or

$H = \left(\begin{matrix}- 2 {e}^{y} & - 2 {e}^{y} x \\ - 2 {e}^{y} x & 2 {e}^{y} + 4 {e}^{y} y + {e}^{y} \left(- {x}^{2} + {y}^{2}\right)\end{matrix}\right)$

so

$H \left(0 , 0\right) = \left(\begin{matrix}- 2 & 0 \\ 0 & 2\end{matrix}\right)$ has eigenvalues $\left\{- 2 , 2\right\}$. This result qualifies point $\left\{0 , 0\right\}$ as a saddle point.

$H \left(0 , - 2\right) = \left(\begin{matrix}- \frac{2}{e} ^ 2 & 0 \\ 0 & - \frac{2}{e} ^ 2\end{matrix}\right)$ has eigenvalues $\left\{- \frac{2}{e} ^ 2 , - \frac{2}{e} ^ 2\right\}$. This result qualifies point $\left\{0 , - 2\right\}$ as a local maximum.

Attached the $f \left(x , y\right)$ contour map near the points of interest