Question

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

Answer 1

Answer:

$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

Let r be the radius of the semicircle, then 2r is the width of the rectangle. Let y be the length of the rectangle. The perimeter of the window consists of two lengths, one width and length of semicircle, then

$p=2y+2r+\pi r.$

Express y:

$y=\dfrac{p-2r-\pi r}{2}.$

The area of the window is

$A=y\cdot 2r+\dfrac{1}{2}\pi r^2.$

Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).