<s>[INST] Find absolute maximum and minimum values by parametrizing the boundaries $f(x,y) = 2\cos x + 3\sin y$  $; R= {(x , y): 0 \leq x \leq 2\pi \\\mbox{and}\\ 0 \leq y \leq \pi} $
I need to find the absolute maximum value and absolute minimum value in the region $R$, and I do have to parametrize the boundary pieces of $R$ to find critical points there.
I tried taking $(x,y) = (r\cos(\theta),r\sin(\theta))$ for $\theta \in [0,2\pi]$
and then I got $h(\theta) = 2\cos(r\cos(\theta))+3\sin(r\sin(\theta))$
After that $h'(\theta)=2r\sin(\theta)\cdot\sin(r\cos(\theta))+3r\cos(\theta)\cdot(\cos(r\sin(\theta))$.
I can't find values of $\theta$ for which $h'(\theta)=0$. 
How should I proceed from here? 
 [/INST] Making the parameterizations
$$
x = \pi(1+\sin (u))\\
y = \frac{\pi}{2}(1+\sin (v))
$$
we have
$$
f(u,v) = 3 \cos \left(\frac{1}{2} \pi  \sin (v)\right)-2 \cos (\pi  \sin (u))
$$
so the stationary points are the solutions for
$$
\left\{
\begin{array}{rcl}
 2 \pi  \cos (u) \sin (\pi  \sin (u))=0 \\
 -\frac{3}{2} \pi  \cos (v) \sin \left(\frac{1}{2} \pi  \sin (v)\right)=0 \\
\end{array}
\right.
$$
giving the solutions
$$
\left[
\begin{array}{ccc}
x & y & f \\
 \pi  & -\frac{\pi }{2} & -2 \\
 \pi  & \frac{\pi }{2} & -2 \\
 \pi  & 0 & 1 \\
 0 & -\frac{\pi }{2} & 2 \\
 0 & \frac{\pi }{2} & 2 \\
 2 \pi  & -\frac{\pi }{2} & 2 \\
 2 \pi  & \frac{\pi }{2} & 2 \\
 0 & 0 & 5 \\
 2 \pi  & 0 & 5 \\
\end{array}
\right]
$$

</s>