(a) Find the coordinates (x,y) which will make the rectangular area A = xy a maximum. (b) What is the value of the maximum area?

(a) Let $(x,y)$ be a point on the ellipse 
\[\frac{x^2}{16}+\frac{y^2}{25}=1.\]
This ellipse can be parametrized as
\[x=4\cos \theta ,  y=5\sin \theta,   0\leq \theta \leq 2\pi.\]
We have
\[A(\theta)=xy=(4\cos \theta)(5\sin \theta)=10(2\sin \theta \cos \theta)=10\sin 2\theta.\]
\[A'(\theta)=20\cos (2\theta)=0    \Rightarrow \cos (2\theta)=0\]
\[2\theta=\frac{\pi}{2} \Rightarrow.  \theta= \frac{\pi}{4}.\]
So the area is maximized if $\theta= \frac{\pi}{4}$ and hence
\[x=4\cos (\frac{\pi}{4})=2\sqrt{2}, y=5\sin(\frac{\pi}{4})=\frac{5\sqrt{2}}{{2}}.\]
So 
\[(x,y)=(2\sqrt{2}, \frac{5\sqrt{2}}{{2}}).\]
(b)
\[ A= xy=2\sqrt{2} \cdot \frac{5\sqrt{2}}{{2}}=10.\]