1
$\begingroup$

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!

  • 0
    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.2012-12-20
  • 0
    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}$2012-12-20
  • 0
    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

1 Answers 1

3

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

  • 0
    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.2012-12-20
  • 0
    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?2012-12-20
  • 0
    '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...2012-12-20
  • 0
    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 $0$.2012-12-20
  • 0
    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 :) ).2012-12-20
  • 0
    Regarding your curiosity about 'curves of critical points', the $y$ axis (which is all made of critical points) in the plot above shifts from being the top of the $\cap$-shaped sections to the bottom of the $\cup$-shaped ones. Can you spot it? Is it graphically clear that you shift from a half line of maximums to a saddle point and then to a half line of minimums? The plot is rotated so that the $x$ axes is parallel to the bottom front left edge...2012-12-20
  • 0
    This clears up one of my other questions: so the critical points $(0,y)$ need not be all of the same nature for different values of $y$?2012-12-20
  • 0
    That's right, your example witnesses all the three types (max, min, saddle) together on the same connected component of critical points. Obviously, each critical point is of exactly one type, but one point does not affect the others (more or less)...2012-12-20
  • 0
    Ok, thanks for your help!2012-12-20
  • 0
    One more question actually: how can you tell that for $y$ greater than $0,$ we have minimums? Did you just evaluate the second derivative and observe this is positive for all y greater than $0$?2012-12-20
  • 0
    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