Find absolute maximum and minimum values by parametrizing the boundaries

Question

$f(x,y) = 2\cos x + 3\sin y$ $; 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?

All you need is that $-1 \le \cos x \le 1$ and $0 \le \sin y \le 1$ on the given intervals. — dxiv, Jun 29 '18 at 23:53
@dxiv I am bit confused because of that r. I kind of get it, but I just can't write it in the form of an answer. — thepanda, Jun 29 '18 at 23:57
$2 \cdot (-1)+3 \cdot 0 \le 2\cos x + 3\sin y \le 2 \cdot 1 + 3 \cdot 1,$ on $,R,$, and both equality cases are attained, so those are the absolute minimum, respectively maximum, values. — dxiv, Jun 30 '18 at 00:01
@dxiv I think the method you are using is without parametric boundaries. The question specifically asked to do it using the parametric method. — thepanda, Jun 30 '18 at 00:16
There is no need (and indeed no point) to parametrize the boundary in this very simple case. But if you have to do it, then see for example Question on parametrization of the boundary of a rectangle in $;\mathbb R^2;$. I tried taking (x,y)=(rcos(θ),rsin(θ)) That's the parametrization of a circle, not a rectangle. — dxiv, Jun 30 '18 at 00:26
@dxiv thanks a lot for pointing me in the correct direction. — thepanda, Jun 30 '18 at 00:30

score 2 · Accepted Answer · answered Jun 30 '18 at 10:52

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

$$ \left[ \begin{array}{ccc} x & y & f \\ \pi & -\frac{\pi }{2} & -2 \\ \pi & \frac{\pi }{2} & -2 \\ \pi & 0 & 1 \\ 0 & -\frac{\pi }{2} & 2 \\ 0 & \frac{\pi }{2} & 2 \\ 2 \pi & -\frac{\pi }{2} & 2 \\ 2 \pi & \frac{\pi }{2} & 2 \\ 0 & 0 & 5 \\ 2 \pi & 0 & 5 \\ \end{array} \right] $$

Find absolute maximum and minimum values by parametrizing the boundaries

1 Answers1