Lisantiom

2022-10-03

Find the critical points of

$f(x,y)={x}^{2}y{e}^{-x-y}.$

$x=0$ 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?

$f(x,y)={x}^{2}y{e}^{-x-y}.$

$x=0$ 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

Beginner2022-10-04Added 6 answers

The set of critical points is $\{(0,y):\text{}y\in \mathbb{R}\}\cup \{(2,1)\}$. The Hessian of $f$ is

$Hf(x,y)=\left[\begin{array}{cc}y{e}^{-x-y}({x}^{2}-4x+2)& x(2-x){e}^{-x-y}(1-y)\\ x(2-x){e}^{-x-y}(1-y)& {x}^{2}{e}^{-x-y}(y-2)\end{array}\right]$

The critical point $(2,1)$ is clearly a local maximum, since $Hf(2,1)$ has deteminant $8{e}^{-6}>0$ and trace $-6{e}^{-3}<0$

This is sufficient to deduce that the points $\{(0,y):\text{}y0\}$ are all local minimums, $\{(0,y):\text{}y0\}$ are local maximums, and $(0,0)$ 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 $f(x,y)={x}^{2}$.

$Hf(x,y)=\left[\begin{array}{cc}y{e}^{-x-y}({x}^{2}-4x+2)& x(2-x){e}^{-x-y}(1-y)\\ x(2-x){e}^{-x-y}(1-y)& {x}^{2}{e}^{-x-y}(y-2)\end{array}\right]$

The critical point $(2,1)$ is clearly a local maximum, since $Hf(2,1)$ has deteminant $8{e}^{-6}>0$ and trace $-6{e}^{-3}<0$

This is sufficient to deduce that the points $\{(0,y):\text{}y0\}$ are all local minimums, $\{(0,y):\text{}y0\}$ are local maximums, and $(0,0)$ 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 $f(x,y)={x}^{2}$.

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