$f(x,y) = 2\cos x + 3\sin y$ $; 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?
Best Answer
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
$$ \left[ \begin{array}{ccc} x & y & f \\ \pi & -\frac{\pi }{2} & -2 \\ \pi & \frac{\pi }{2} & -2 \\ \pi & 0 & 1 \\ 0 & -\frac{\pi }{2} & 2 \\ 0 & \frac{\pi }{2} & 2 \\ 2 \pi & -\frac{\pi }{2} & 2 \\ 2 \pi & \frac{\pi }{2} & 2 \\ 0 & 0 & 5 \\ 2 \pi & 0 & 5 \\ \end{array} \right] $$