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?


  • $\begingroup$ 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. $\endgroup$ – CAF Dec 20 '12 at 13:43
  • $\begingroup$ For $x=0,$ $y$ can be any value. $f(0,y) = f(y,0) = 0\, \forall \,y,x$. But $f(x,x) = x^3 e^{-2x}$ $\endgroup$ – CAF Dec 20 '12 at 14:13
  • $\begingroup$ 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? $\endgroup$ – 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.

Plotted with <code>gnuplot</code>, ranging in the square $[-0.1,0.1]\times[-0.1,0.1]$ 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 $f(x,y)=x^2$.

  • $\begingroup$ 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. $\endgroup$ – AndreasT Dec 20 '12 at 14:50
  • $\begingroup$ Many thanks for your response. What is meant by a 'singular' point? Also, it would appear from Wolfram Alpha that the point (0,0) is not a saddle point? $\endgroup$ – CAF Dec 20 '12 at 15:02
  • $\begingroup$ '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... $\endgroup$ – AndreasT Dec 20 '12 at 15:19
  • $\begingroup$ 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 $0$. $\endgroup$ – CAF Dec 20 '12 at 15:28
  • $\begingroup$ 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 :) ). $\endgroup$ – AndreasT Dec 20 '12 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.