I'm trying to find stationary points for $f(x,y)=x^2e^{-y}$ so I took partial derivatives to find the gradient vector and got $\langle 2xe^{-y} , -x^2 e^{-y}\rangle$

Now I'm trying to find at what points the gradient vector becomes vector $<0,0>$ so I made the equations: $2xe^{-y}=0 ; -x^2 e^{-y}=0$

From the first one I got that $x=0$. But if I plug that in the second equation, then $0* e^{-y}=0$ which makes me think $y$ could have just any value (let's call it $n$).

Is that correct? If so, then stationary points would be of the form $<0, n>$ which confuse me a bit on how to calculate the hessian matrix.

As for the hessian matrix, I got that:





Which leaves me with $2x^2e^{-2y}-4x^2e^{-2y}$

So if $y=n$ (any number) how can I evaluate the hessian at this generic point?

  • 1
    $\begingroup$ The hessian is always zero in that case (x=0). $\endgroup$ – MattG88 Dec 3 '16 at 18:14
  • $\begingroup$ So my reasoning is correct then, and I should evaluate the hessian using y as a generic value? (I know in this case it'll be 0 because x=0, but what about other cases where I have a generic value?) $\endgroup$ – Floella Dec 3 '16 at 18:17
  • 1
    $\begingroup$ Yes it is right, y is generic. $\endgroup$ – MattG88 Dec 3 '16 at 18:23

Your function is always positive or equal zero. The hessian is zero so you can't say if $(0,y)$ are max or min points; but if you evaluate the function in those points you get zero, so $(0,y)$ must be minimum points, because $f(x,y)\ge0$.

Along the y-axis we have infinite minimum points.

You could say at first look that $(0,y)$ were minimum points without making any calculation just seeing the function, so I advise you to check the function before doing gradient and so on.


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.