Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 30

A Norman window has the shape of a rectangle surrounded by a semicircle. If the perimeter of the window is 30ft, find the dimensions of the window so that the greatest possible amount of light is admitted.




Let $P$ and $S$ be the perimeter and surface area of the window.

$
\begin{equation}
\begin{aligned}
P &= x + 2y + \frac{2\pi r}{2} = x + 2y + \pi r\\
\\
\text{since } r &= \frac{x}{2}\\
\\
P &= x + 2y + \pi \frac{x}{2} = x + x \frac{\pi}{2} + 2y
\end{aligned}
\end{equation}
$


Also, the Surface Area of the window is...

$
\begin{equation}
\begin{aligned}
s &= xy + \frac{\pi r^2}{2}; \quad r = \frac{x}{2}\\
\\
s &= xy + \frac{\pi \left( \frac{x}{2} \right)^2}{2} = xy + \frac{\pi x^2}{8}
\end{aligned}
\end{equation}
$


We have, $\displaystyle P = x + x \frac{\pi}{2} + 2y = 30$

Solving for $y$

$
\begin{equation}
\begin{aligned}
2y &= 30 - \left( \frac{2x + x \pi}{2} \right)\\
\\
y &= \frac{30 - \left( \frac{2x + x \pi}{2} \right) }{2} = \frac{60-2x+x\pi}{4}
\end{aligned}
\end{equation}
$


Substituting the value of $y$ to the equation of the Surface Area...
$\displaystyle s = x \left( \frac{60-2x-\pi x}{4} \right) + \frac{\pi x^2}{8} = \frac{60x-2x^2-\pi x^2}{4} + \frac{\pi x ^2}{8}$
Taking the derivative with respect to $x$, we have...

$
\begin{equation}
\begin{aligned}
s' &= \frac{60-4x - 2 \pi x}{4} + \frac{2 \pi x}{8}\\
\\
\text{when } s' &= 0 \\
\\
0 &= \frac{60-4x-2\pi x}{4} + \frac{2\pi x}{8}\\
\\
0 &= \frac{120-8x-4 \pi x + 2 \pi x}{8}\\
\\
0 &= 120-8x - 4\pi x + 2 \pi x\\
\\
0 &= 120 - 8x - 2 \pi x\\
\\
2 \pi x + 8x &= 120\\
\\
2x(\pi + 4) &= 120\\
\\
x &= \frac{60}{(\pi + 4)}\text{ft}
\end{aligned}
\end{equation}
$

so when, $\displaystyle x = \frac{60}{(\pi + 4)}$
$\displaystyle r = \frac{x}{2} = \frac{ \left( \frac{60}{\pi + 4}\right)}{2} = \frac{30}{(\pi + 4)}$ft

and,

$
\begin{equation}
\begin{aligned}
y &= \frac{60 - 2 \left( \frac{60}{\pi + 4} \right) - \left( \frac{60}{\pi + 4} \right) \pi}{4}\\
\\
y &= 4.20 \text{ft}
\end{aligned}
\end{equation}
$


Therefore, the greatest possible amount of height will be admitted if the area of the window is...

$
\begin{equation}
\begin{aligned}
s &= xy + \frac{\pi x^2}{8}\\
\\
s &= \left( \frac{60}{(\pi + 4)} \right) (4.20) + \frac{\pi \left( \frac{60}{\pi + 4} \right)^2}{8}\\
\\

\end{aligned}
\end{equation}\\
\boxed{s = 63\text{ft}^2}
$