IMC 1994 · Day 2 · P8

We have $f(1,0) = e^{-1}$, $f(0,1) = -e^{-1}$ and $t e^{-t} \le 2 e^{-2}$ for $t \ge 2$. Therefore $|f(x,y)| \le (x^2 + y^2) e^{-x^2 - y^2} \le 2 e^{-2} < e^{-1}$ for $(x,y) \notin M = \{ (u,v) : u^2 + v^2 \le 2 \}$ and $f$ cannot attain its minimum and its maximum outside $M$. Part a) follows from the compactness of $M$ and the continuity of $f$. Let $(x,y)$ be a point from part b). From $\frac{\partial f}{\partial x}(x,y) = 2x (1 - x^2 + y^2) e^{-x^2 - y^2}$ we get $$\tag{1} x (1 - x^2 + y^2) = 0.$$ Similarly $$\tag{2} y (1 + x^2 - y^2) = 0.$$ All solutions $(x,y)$ of the system (1), (2) are $(0,0)$, $(0,1)$, $(0,-1)$, $(1,0)$ and $(-1,0)$. One has $f(1,0) = f(-1,0) = e^{-1}$ and $f$ has global maximum at the points $(1,0)$ and $(-1,0)$. One has $f(0,1) = f(0,-1) = -e^{-1}$ and $f$ has global minimum at the points $(0,1)$ and $(0,-1)$. The point $(0,0)$ is not an extrema point because of $f(x,0) = x^2 e^{-x^2} > 0$ if $x \ne 0$ and $f(y,0) = -y^2 e^{-y^2} < 0$ if $y \ne 0$.