Finding the maximum and minimum points of $f(x,y)=\frac{2\sin(y)}{1+(x-\pi)^2}$ over the region $R=\{(x,y) \in \mathbb{R}^2 : 0 \le x \le 2\pi, \ 0 \le y \le \pi\}$.
I have found $f_x=\frac{-4(x-\pi)\sin(y)}{(1+(x-\pi)^2)^2}$ and $f_y=\frac{2\cos(y)}{1+(x-\pi)^2}$
For $f_x=0$ it should be that $-4(x-\pi)\sin(y)=0$ so $x=\pi$ or $y=0, \pi$
For $f_y=0$, $\ 2\cos(y)=0$ so $y=\frac{\pi}{2}$
But how do I find the points themselves? And how do I evaluate the function in the "borders" delimited by $R$?
Thanks for your time!