Question

A window consists of a rectangle surmounted by semi-circle. What shape gives the most area for a given perimeter?

Answer 1

Let us skerch a picture of such a window:

Let $a$ and $b$ be lengths of a corresponding rectangle. Then the surmounted semi-circle is of radius $\frac{a}{2}$.

The perimeter of the figure is $P=2b+a+\frac{\pi a}{2}=2b+\frac{2+\pi}{2}a$ and the area is $S=ab+\frac{1}{2}\pi(\frac{a}{2})^2=ab+\frac{\pi a^2}{8}.$ It follows that $b=\frac{P}{2}-\frac{2+\pi}{4}a$ and therefore, $S=a(\frac{P}{2}-\frac{2+\pi}{4}a)+\frac{\pi a^2}{8}=\frac{P}{2}a-\frac{2+\pi}{4}a^2+\frac{\pi a^2}{8} =\frac{P}{2}a-\frac{4+\pi}{8}a^2$. Since the perimeter is constant, $S'=\frac{P}{2}-\frac{4+\pi}{4}a$. If $S'=0,$ then $\frac{P}{2}-\frac{4+\pi}{4}a=0$, and thus $a=\frac{2P}{4+\pi}$. Taking into account that $S''=-\frac{4+\pi}{4}<0$, we conclude that $a=\frac{2P}{4+\pi}$ is a point of maximum of $S$. It follows that $b=\frac{P}{2}-\frac{2+\pi}{4}\frac{2P}{4+\pi}=\frac{P}{2}-\frac{(2+\pi)P}{2(4+\pi)}=\frac{4P+\pi P-2P-\pi P}{2(4+\pi)}=\frac{P}{4+\pi}$, and thus $a=2b$. We conclude that for $b=\frac{P}{4+\pi}$ and $a=2b=\frac{2P}{4+\pi}$ the area is maximum.