Lisantiom

2022-10-03

Find the critical points of
$f\left(x,y\right)={x}^{2}y{e}^{-x-y}.$
$x=0$ and $x=2$ both satisfy ${f}_{x}=0,{f}_{y}=0$, and when $x=2,y=1$. But when $x=0,y$,$x=0,y$ is arbritary. So how can I determine whether the critical point at $0$ is max/min/saddle?

Kelly Ibarra

The set of critical points is . The Hessian of $f$ is
$Hf\left(x,y\right)=\left[\begin{array}{cc}y{e}^{-x-y}\left({x}^{2}-4x+2\right)& x\left(2-x\right){e}^{-x-y}\left(1-y\right)\\ x\left(2-x\right){e}^{-x-y}\left(1-y\right)& {x}^{2}{e}^{-x-y}\left(y-2\right)\end{array}\right]$
The critical point $\left(2,1\right)$ is clearly a local maximum, since $Hf\left(2,1\right)$ has deteminant $8{e}^{-6}>0$ and trace $-6{e}^{-3}<0$
This is sufficient to deduce that the points are all local minimums, are local maximums, and $\left(0,0\right)$ is a saddle.

If you wonder what a curve of critical points graphically means, think of a half pipe placed horizontally on the ground: the set of points that touch the ground is all made of local (global too...) minimums. For a more mathematical example, just consider the shape of the surface $f\left(x,y\right)={x}^{2}$.

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