Find the extreme values of $f(x,y)=xy$ on $D=\{(x,y)|1 \leq x^2+y^2 \leq 4\}$

Question

This would have to be done using conditional extremes(Lagrange method), and maybe some topological properties.I do not know how to do this, I have only done cases where the $D$ would be defined with one inequality.

Answer 1

The map $$ \left(x,y\right)\mapsto \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right) $$ is an isometry of $\mathbb{R}^2$, hence your problem is equivalent to finding the extreme values of $\frac{u^2-v^2}{2}$ over $1\leq u^2+v^2\leq 4$. They clearly are $\color{red}{\pm 2}$, and they clearly occur at points of the outer boundary, also because $f(x,y)=xy$ is a harmonic function ($\Delta f=0$).

Answer 2

First of all, what does $D$ look like? It is a circle with radius $2$ where a circle with radius $1$ is cut out.
In this particular problem, you can first find the maximum of $f(x,y)$ on the circle with radius $1$, which is $D = \{ (x,y) \mid x^2+y^2 \le 1\}$, so you have one inequality, thus you should be able to do so.
Then, you can find the maximum of $f(x,y)$ on the circle with radius $2$, and if this maximum is strictly bigger than the maximum on the first circle, you see that the maximum on $D$ is the latter.

An other way to do this, is switching to polar coordinates.
Then $f(r, \theta) = r^2 \cos( \theta) \sin(\theta)$ with $D = \{ (r ,\theta) \mid 1 \le r^2 \le 4, 0 \le \theta \le 2 \pi \}$.
Then you see that $f(r ,\theta)$ attains it maximum when $r=1$ in case $\cos( \theta) \sin(\theta)$ is always negative, or $r=2$ when $\cos( \theta) \sin(\theta)$ attains positive values. And in both cases, $\cos( \theta) \sin(\theta)$ can be easily maximized for $\theta$.