# Critical points

• Feb 7th 2011, 11:52 AM
Killer
Critical points
I have to find the critical points of $\displaystyle f(x,y)=xye^{-x^2-y^2}$ and decide if they're local maxima or local minima or neither of both. (This one has a name but I don't remember it.)

Well first we have that $\displaystyle \nabla f(x,y)=\left( \left( y-2{{x}^{2}}y \right){{e}^{-{{x}^{2}}-{{y}^{2}}}},\left( x-2x{{y}^{2}} \right){{e}^{-{{x}^{2}}-{{y}^{2}}}} \right),$ so we have $\displaystyle y(1-2x^2)=0$ and $\displaystyle x(1-2y^2)=0.$ Critical points are

\displaystyle \begin{aligned} x=0&,y=0 \\ x=-\frac{1}{\sqrt{2}}&,y=-\frac{1}{\sqrt{2}} \\ x=-\frac{1}{\sqrt{2}}&,y=\frac{1}{\sqrt{2}} \\ x=\frac{1}{\sqrt{2}}&,y=-\frac{1}{\sqrt{2}} \\ x=\frac{1}{\sqrt{2}}&,y=\frac{1}{\sqrt{2}}. \end{aligned}

Now the Hessian is

$\displaystyle H\,f(x,y)=\left[ \begin{matrix} 2xy\left( 2{{x}^{2}}-3 \right){{e}^{-{{x}^{2}}-{{y}^{2}}}} & \left( 2{{x}^{2}}-1 \right)\left( 2{{y}^{2}}-1 \right){{e}^{-{{x}^{2}}-{{y}^{2}}}} \\ \left( 2{{x}^{2}}-1 \right)\left( 2{{y}^{2}}-1 \right){{e}^{-{{x}^{2}}-{{y}^{2}}}} & 2xy\left( 2{{y}^{2}}-3 \right){{e}^{-{{x}^{2}}-{{y}^{2}}}} \\ \end{matrix} \right].$

Now my problem is when a point is a maximum or minimum, what are the criteria?
• Feb 7th 2011, 11:59 AM
alexmahone