Answer
(a) $A(x) = -(\frac{\pi+4}{8})~(x-\frac{60}{\pi+4})^2+\frac{4500}{\pi+4}$
(b) The width of the window is $~~\frac{60}{\pi+4}~ft \approx 8.40~ft$
The height of the rectangular part of the window is $~~\frac{30}{\pi+4}~ft \approx 4.20~ft$
Work Step by Step
(a) Let $x$ be the width of the rectangle. Then $\frac{x}{2}$ is the radius of the semicircle.
Let $y$ be the height of the rectangle.
We can use the perimeter to find an expression for $y$:
$P = x+2y+(\frac{1}{2}) (2\pi) (\frac{x}{2}) = 30$
$2y = 30-x-\frac{\pi~x}{2}$
$y = 15-\frac{x}{2}-\frac{\pi~x}{4}$
We can find an expression for the area:
$A = xy+\frac{1}{2}~\pi (\frac{x}{2})^2$
$A = (x)(15-\frac{x}{2}-\frac{\pi~x}{4})+\frac{\pi~x^2}{8}$
$A = 15x-\frac{x^2}{2}-\frac{\pi~x^2}{4}+\frac{\pi~x^2}{8}$
$A = 15x-\frac{x^2}{2}-\frac{\pi~x^2}{8}$
$A = 15x-\frac{1}{2}-\frac{\pi}{8}~x^2$
$A = 15x-(\frac{\pi+4}{8})~x^2$
$A = -(\frac{\pi+4}{8})~(x^2-\frac{120}{\pi+4}~x)$
$A = -(\frac{\pi+4}{8})~[x^2-\frac{120}{\pi+4}~x+\frac{3600}{(\pi+4)^2}]+(\frac{\pi+4}{8})[\frac{3600}{(\pi+4)^2}]$
$A = -(\frac{\pi+4}{8})~[x^2-\frac{120}{\pi+4}~x+\frac{3600}{(\pi+4)^2}]+\frac{4500}{\pi+4}$
$A = -(\frac{\pi+4}{8})~(x-\frac{60}{\pi+4})^2+\frac{4500}{\pi+4}$
This function models the area of the window.
(b) We can see that this parabola has a vertex at the point $(\frac{60}{\pi+4}, \frac{4500}{\pi+4})$
The maximum area is $\frac{4500}{\pi+4}$ when the width of the window is $~~\frac{60}{\pi+4}$
We can find the height of the rectangle:
$y = 15-\frac{x}{2}-\frac{\pi~x}{4}$
$y = 15-\frac{\frac{60}{\pi+4}}{2}-\frac{(\pi)~(\frac{60}{\pi+4})}{4}$
$y = 15-\frac{30}{\pi+4}-\frac{15 \pi}{\pi+4}$
$y = 15-\frac{15 \pi+30}{\pi+4}$
$y = \frac{15 \pi+60}{\pi+4}-\frac{15 \pi+30}{\pi+4}$
$y = \frac{30}{\pi+4}$
The width of the window is $~~\frac{60}{\pi+4}~ft \approx 8.40~ft$
The height of the rectangular part of the window is $~~\frac{30}{\pi+4}~ft \approx 4.20~ft$
