A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 33 feet?

Answer 1

Let x = the width of the rectangle.

The perimeter of the window is 33 feet, so:
2x + πx = 33

Solving for x, we get:
x = (33 - π)/2

The area of the window is the area of the rectangle plus the area of the semicircle:
A = x(x + 2πx/2)
= x^2 + πx^2/2
= (33 - π)^2/4 + (π/2)(33 - π)^2/4
= (33^2 - 2π33 + π^2)/4 + (π/2)(33^2 - 2π33 + π^2)/4
= (33^2 - 2π33 + π^2 + (π/2)(33^2 - 2π33 + π^2))/4
= (33^2 - 2π33 + (3π^2 + 33π)/2)/4
= (33^2 + 33π + 3π^2)/4
= (33 + 3π)(33 + π)/4
= 1089π/4

Therefore, the area of the largest possible Norman window with a perimeter of 33 feet is 1089π/4 square feet.