# How do you maximize a window that consists of an open rectangle topped by a semicircle and is to have a perimeter of 288 inches?

Jul 1, 2015

I found two values, for the width and height (of the rectangular part) of your window:
$80.6$ in and $40.4$ in

#### Explanation:

Perimeter is:
$P = 2 a + b + \pi \left(\frac{b}{2}\right) = 288$
so: $a = \frac{1}{2} \left[288 - b - \frac{\pi}{2} b\right]$ (1)
Area is:
$A = b \cdot a + \frac{\pi}{2} {\left(\frac{b}{2}\right)}^{2}$ using the value of $a$ from (1):
$A = b \frac{1}{2} \left[288 - b - \frac{\pi}{2} b\right] + \frac{\pi}{2} {\left(\frac{b}{2}\right)}^{2} =$
$= 144 b - {b}^{2} / 2 - \frac{\pi}{4} {b}^{2} + \frac{\pi}{8} {b}^{2} = 144 b - {b}^{2} \left(\frac{1}{2} + \frac{\pi}{4} - \frac{\pi}{8}\right)$
Maximize the area deriving it and setting it equal to zero:
$A ' = 144 - 2 b \left(\frac{1}{2} + \frac{\pi}{4} - \frac{\pi}{8}\right) = 0$
so that $b = 80.6$in
So from (1): $a = 40.4$in

Where I used for the semicircle:
Perimeter$= \frac{1}{2} \cdot 2 \pi r$
Area$= \frac{1}{2} \cdot \pi {r}^{2}$