Maximum of the function $f(x, y)=(x^2+y^2)e^{-x^2-y^2}$.

Question

Let $f:\mathbb{R^2}\rightarrow \mathbb{R}$ be the function defined by $f(x, y)=(x^2+y^2)e^{-x^2-y^2}$. Then

1) the point $(0 ,0)$ is a global minimum for the function $f$.

2) the function $f$ does not have maximum.

3) the function $f$ attains its maximum at a point in $\{ (x,y)\in \mathbb{R^2}:x^2+y^2=1\}$

4) the function $f$ has saddle point in $\{ (x,y)\in \mathbb{R^2}:x^2+y^2 \neq 1\}$

I use transformation $x=r \cos \theta$ and $y=r \sin \theta$ then $f(r)=r^2e^{-r^2}$. I find that $f$ attains maximum at $r=1$ . So the option (3) is correct and other are false. Is my analysis correct?

Answer 1

HINT: Write $u=x^2+y^2$. Write $g(u)= ue^{-u}$. To answer the question in the title of your thread: Find the value of $u$ s.t. $g(u)$ is maximized [and in general look at the behaviour of $g(u); u \ge 0$]. Translate this back to the behaviour of $f(x,y)$ as $u=x^2+y^2$ varies.

[As far as the remaining questions: Suppose you have a saddle point of $f$ at $(x_0,y_0)$. Then what is the behaviour of $g(u_0)$ at $u_0=x_0^2+y_0^2$? Also note that by symmetry, iff $f$ has a saddle point at $(x_0,y_0)$ it also has a saddle point at $\left(\sqrt{x^2_0+y^2_0}, 0\right)$ and for that matter anywhere else on the circle $x^2+y^2=u_0$ where $u_0=x^2_0+y^2_0$...make sure you see why]

Answer 2

If $3)$ is true, then $2)$ is necessarily false. It's also obvious that $f$ attains minimum at $(0,0)$, as $x^2+y^2 \ge 0$ and $e^{-x^2-y^2} > 0$. Thus $1)$ is true.

Anyway to find the critical points, compute the partial derivatives and check where they are zero. We have:

$$f_x = 2xe^{-x^2-y^2} - 2x(x^2+y^2)e^{-x^2-y^2} = 2xe^{-x^2-y^2}(1-x^2-y^2)$$ $$f_y = 2ye^{-x^2-y^2}(1-x^2-y^2)$$

So for the critical points we must have $x=0,y=0$ or $x^2+y^2=1$. Thus as at $(0,0)$ we have local minimum we conclude that there isn't a saddle point outside of $x^2+y^2=1$. Hence $4)$ is false.