Question

A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimensions of the rectangular part of a Norman window to allow in as much light as possible, if there are only 12 ft of the frame material available?

Answer 1

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

$A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}$

The restriction is that the perimeter have to be 12 ft at most:

$P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12$

We can express b in function of a as:

$2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a$

Then, the area become:

$A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2$

To maximize the area, we derive and equal to zero:

$\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36$

Then, b is:

$b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68$