Answer to Question #274597 in Calculus for Artika

Question

A church window consisting of a rectangle topped by a semicircle is to have a perimeter p. Find the radius of the semicircle if the area of the window is to be a maximum.

Answer 1

The rectangle has dimensions: $2r$ and $x$.

Tne semicircle: $\frac{1}{2}\cdot 2\cdot \pi \cdot r=\pi \cdot r$ .

$p=2r+2x+\pi \cdot r\\ x=\frac{1}{2}(p-r(\pi+2))$

The area of the window is to be a maximum:

$S=x\cdot 2r+\frac{1}{2}\pi \cdot r^2=\\ =(p-r(\pi+2))\cdot r+\frac{1}{2}\pi \cdot r^2=\\ =pr-r^2(\pi+2)+\frac{1}{2}\pi \cdot r^2\\ S'=p-2r(\pi+2)+\pi r=p-\pi r-4r\\ p-\pi r-4r=0\\ r(\pi+4)=p\\ r=\frac{p}{\pi+4}$

If $r\in (-\infty, \frac{p}{\pi+4}), S'>0$ .

If $r\in (\frac{p}{\pi+4},\infty), S'<0$ .

$S(\frac{p}{\pi+4})$ - max.

$r=\frac{p}{\pi+4}$ .