Conceptual question about a critical point 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!
 A: 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.

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 $f(x,y)=x^2$.
