Max and min value of $f(x,y)$ $$f(x,y) = 2\cdot x\cdot y$$
$$x^2+y^2 \leq 4$$
I have no idea about this question.There is a region.How can I solve?

Since $(|x|-|y|)^2\geq 0$, we have
$$
2|xy|\leq x^2+y^2\leq 4.
$$
Hence
$$
-4\leq 2xy\leq 4. 
$$
If $x=y=\sqrt{2}$ we have $x^2+y^2=4\leq 4$ and $2xy=4$.
If $x=\sqrt{2}, y=-\sqrt{2}$ we have $x^2+y^2=4\leq 4$ and $2xy=-4$.
Therefore, the minimum value of $2xy$ is $-4$ and the maximum value of $2xy$ is $4$.
