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Say I want to find the critical points of $f(x,y) = x^2 ye^{-x-y}.$ $x=0$ and $x=2$ both satisfy $f_x = 0, f_y = 0$, and when $x=2, y=1$. But when $x=0, y$ is arbritary. So how can I determine whether the critical point at $0$ is max/min/saddle?

Also, geometrically, what does this mean?

Thanks!

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    So from the surface you gave, for $x=0, y$ can be any value and $f(x,y)$ does not increase or decrease so it is level at $z=0$. What does this tell us about the nature of the point? Can I just substitute an arbritary value for $y$ into the Hessian and see what I get?2012-12-20

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The set of critical points is $\{(0,y):~y\in\mathbb{R}\}\cup\{(2,1)\}$. The Hessian of $f$ is $ Hf(x,y) = \begin{bmatrix} ye^{-x-y}(x^2-4x+2) & x(2-x)e^{-x-y}(1-y) \\ x(2-x)e^{-x-y}(1-y) & x^2e^{-x-y}(y-2) \end{bmatrix} $ The critical point $(2,1)$ is clearly a local maximum, since $Hf(2,1)$ has deteminant $8e^{-6}>0$ and trace $-6e^{-3}<0$.

For the other critical points the Hessian is of no use, since it is singular. Therefore we have to find a way around: first of all let us notice that $f(0,y)=0$ for all $y$. It is trivial to notice that \begin{align*} x^2ye^{-x-y}\geq 0 & \quad\text{for all $x$ and for all $y>0$} \\ x^2ye^{-x-y}\leq 0 & \quad\text{for all $x$ and for all $y<0$} \end{align*} This is sufficient to deduce that the points $\{(0,y):~y>0\}$ are all local minimums, $\{(0,y):~y<0\}$ are local maximums, and $(0,0)$ is a saddle.

Plotted with <code>gnuplot</code>, ranging in the square <span class=[-0.1,0.1]\times[-0.1,0.1] to show the nature of $(0,0)$.">

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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    The Hessian has nothing to do with that, actually nor does the gradient. The reason is simply algebraic: as $x^2ye^{-x-y}\geq 0$ for y> 0, and is equal to $0$ in $x=0$, then the function at every point near $(x,y)$ is higher (or at least has the same height $0$) than its value at $(0,y)$. Specifically, every point $(0,y)$ with $y\geq 0$ is a **global minimum** in the half plane $\{(x,y):~y\geq 0\}$, and the converse apply to the other half plane.2012-12-20