# 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!

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I know about the 2nd derivative test, but then I have to evaluate the Hessian at the critical point $(x_o,y_o)$. I have $x_o$, but as I said, $y$ is left arbritary. – CAF Dec 20 '12 at 13:43
For $x=0,$ $y$ can be any value. $f(0,y) = f(y,0) = 0\, \forall \,y,x$. But $f(x,x) = x^3 e^{-2x}$ – CAF Dec 20 '12 at 14:13
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? – CAF Dec 20 '12 at 14:24

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 $f(x,y)=x^2$.
In simpler cases, when the Hessian is not singular, you just have to solve $$\begin{cases} {\rm det}Hf(x,y)>0 \\ (x,y) \in\{ \text{critical points}\} \end{cases}$$ The solutions are either minimums or maximums, depending on the trace of $Hf$, while the other critical points are saddles. – AndreasT Dec 20 '12 at 14:50
'Singular' is referred to the Hessian matrix (which is the second derivatives matrix $Hf=\begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{bmatrix}$. A matrix is said to be singular if its determinant is $0$, i.e. if it is not invertible. Concerning the point $(0,0)$, I plotted the function with gnuplot and it definitely look like a saddle. I will try to add the output to my answer... – AndreasT Dec 20 '12 at 15:19
That's nice. What about the fact that when I got $x=0$ to be a critical point, $y$ could be arbritary? From a plot in Wolfram, it appears that at $x=0, y$ can be any value and $z=0$. Does this mean all points of the form $(0,y)$ are critical?For fun, I put in $y=1$, and I got a nonzero output for the Hessian? (Different for the outcome when $y=0$, in which case, as you said, the Hessian is $0$. – CAF Dec 20 '12 at 15:28
I am not getting your first question... Points in $\mathbb{R}^2$ are of the form $(x,y)$ so $x=0$ is not a point, rather a (equation of a) line. Concerning your second question, yes, you are right, all points of the form $(0,y)$ are critical. Remember that critical points are only those on which the gradient is null, and only on those studying the Hessian is relevant for the matter (or, then again, I did not get your 'For fun...' statement, sorry :) ). – AndreasT Dec 20 '12 at 15:52