# An arched window formed from the union of a semicircle and a rectangle has total perimeter 540"cm". What is the maximum possible area of the window?

Jun 27, 2015

Maximum area is achieved when the bounding rectangle is a square of side $\frac{1080}{\pi + 4} c m$

#### Explanation:

Let the length of the base of the rectangle be $t$ and its height $h$.

Then the length of the perimeter of the semicircle is $\frac{\pi}{2} t$

The total perimeter of the window is $\frac{\pi}{2} t + t + 2 h = 540 c m$

So $2 h = 540 c m - \left(\frac{\pi}{2} + 1\right) t$

So $h = 270 c m - \left(\frac{\pi}{4} + \frac{1}{2}\right) t$

The area of the rectangle is $t \cdot h = t \cdot \left(270 c m - \left(\frac{\pi}{4} + \frac{1}{2}\right) t\right)$

The area of the semicircle is $\frac{1}{2} \pi {\left(\frac{t}{2}\right)}^{2} = \frac{\pi}{8} {t}^{2}$

So the total area of the window is

$t \cdot \left(270 c m - \left(\frac{\pi}{4} + \frac{1}{2}\right) t\right) + \frac{\pi}{8} {t}^{2}$

$= \left(\frac{\pi}{8} - \left(\frac{\pi}{4} + \frac{1}{2}\right)\right) {t}^{2} + t \cdot 270 c m$

$= t \left(270 c m - \left(\frac{\pi}{8} + \frac{1}{2}\right) t\right)$

This has maximum value when $t = \frac{135}{\frac{\pi}{8} + \frac{1}{2}} c m = \frac{1080}{\pi + 4} c m$

$h = 270 c m - \left(\frac{\pi}{4} + \frac{1}{2}\right) t = 270 c m - \frac{1080 \left(\frac{\pi}{4} + \frac{1}{2}\right)}{\pi + 4} c m$

$= 270 c m - 270 \frac{\pi + 2}{\pi + 4} c m$

$= 270 \frac{\left(\pi + 4\right) - \left(\pi + 2\right)}{\pi + 4} c m$

$= 270 \left(\frac{2}{\pi + 4}\right) c m$

$= \frac{540}{\pi + 4} c m = \frac{t}{2}$

The total height of the window is $h + \frac{t}{2} = t$

Basically the maximum area is achieved when the bounding rectangle is a square of side $\frac{1080}{\pi + 4} c m$