Question

Find the absolute maximum of f(x,y) = e^(-x^2-y^2)(x^2+2y^2) on x^2 + y^2 < 2

Answer 1

$f(x,y)=e^{-x^2-y^2}(x^2+y^2)$

Notice that converting to polar coordinates, setting
$x=r\cos\theta$
$y=r\sin\theta$
$\implies r^2=x^2+y^2$

allows us to consider $f(x,y)$ as a function of one variable; let's call it $F(r)$, where

$f(x,y)\equiv F(r)=re^{-r}$

Then

$F'(r)=e^{-r}(1-r)=0\implies r=1$

We have $F'(r)>0$ for $r<1$, and $F'(r)<0$ for $r>1$, which means $F$ is increasing, then decreasing as $r$ exceeds 1. This suggests that extrema occur for $f(x,y)$ wherever $r^2=x^2+y^2=1$, i.e. along the intersection of the cylinder $x^2+y^2=1$ and $f(x,y)$.

Computing the second derivative of $F(r)$ and setting equal to 0 gives

$F''(r)=-e^{-r}(2-r)=0\implies r=2$

as a possible point of inflection. We have $F''(r)<0$ for $r<2$, and namely when $r=1$, which means $F(r)$ is concave downward around this point. This confirms that $r=1$ is a site of a maximum. Along this path, we have a maximum value of $F(1)=e^{-1}\approx0.368$.

Next, to check for possible extrema along the border, we can parameterize $f(x,y)$ by $x=\sqrt2\cos t$ and $y=\sqrt2\sin t$, so that

$x^2+y^2=(\sqrt2\cos t)^2+(\sqrt2\sin t)^2=2$

and we can think of $f(x,y)$ as a function a single variable, $F(t)$, where

$F(t)=2e^{-2}\approx0.271$

In other words, $f(x,y)$ is constant along its boundary $x^2+y^2=2$, and this is smaller than the maximum we found before.

So to recap, the maximum value of $f(x,y)$ is $\dfrac1e\approx0.368$, which is attained along the surface above the circle $x^2+y^2=1$ in the $x-y$ plane.