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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:

$f_{xx}(x,y)=2e^{-y}$

$f_{xy}(x,y)=-2xe^{-y}$

$f_{yy}(x,y)=x^2e^{-y}$

$f_{yx}(x,y)=-2xe^{-y}$

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?

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    $\begingroup$ 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
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    $\begingroup$ Yes it is right, y is generic. $\endgroup$ – MattG88 Dec 3 '16 at 18:23
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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.

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