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

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

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+(\pi r^2)/(2)=ab+(\pi)/(2)\left ((a)/(2)\right)^2=ab+(\pi a^2)/(8)`

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

`P=(a+2b)+(\pi a)/(2)=2b+a+((\pi)/(2)) a=2b+(1+(\pi)/(2))a=12`

We can express b in function of a as:

`2b+(1+(\pi)/(2))a=12\\\\\\2b=12-(1+(\pi)/(2))a\\\\\\b=6-\left((1)/(2)+(\pi)/(4)\right)a`

Then, the area become:

`A=ab+(\pi a^2)/(8)=a(6-\left((1)/(2)+(\pi)/(4)\right)a)+(\pi a^2)/(8)\\\\\\A=6a-\left((1)/(2)+(\pi)/(4)\right)a^2+(\pi a^2)/(8)\\\\\\A=6a-\left((1)/(2)+(\pi)/(4)-(\pi)/(8)\right)a^2\\\\\\A=6a-\left((1)/(2)+(\pi)/(8)\right)a^2`

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

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

Then, b is:

`b=6-\left((1)/(2)+(\pi)/(4)\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68`