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...

$\begin{equation} \begin{aligned} s &= xy + \frac{\pi x^2}{8}\\ \\ s &= \left( \frac{60}{(\pi + 4)} \right) (4.20) + \frac{\pi \left( \frac{60}{\pi + 4} \right)^2}{8}\\ \\ \end{aligned} \end{equation}\\ \boxed{s = 63\text{ft}^2}$