Question

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 31 feet?

Answer 1

Let r be the radius of the semi-circle and h be the height of the rectangular portion. Thus Width of the window = 2r.

Now perimeter of this window ( P ) = πr + 2 ( h + 2 r )

P = π r + 2 h + 4 r

P = 2 h + r ( π +4 )

2 h + r ( π +4 ) = 31

=> h = (1/2) [ 31 - r (π + 4 ) ]………………… (i)

If A in² is the area of the window then ---

A = (1/2) πr² + 2 r h ……………………………. (ii)

Substituting the value of h from Eqn. (i) in eqn. (ii) we get –

A = 31 r - r² [ ( π/2 ) + 4 ] …………………….. Note it very carefully.

For finding the maximum value of A , dA/dr = 0

dA/dr = 31 - 2 r [ ( π/2 ) + 4 ] = 0

r = 31 / [ π + 8 ] Hence h = 62 / (π + 8)

Hence the largest area

A(max) = (1/2) πr² + 2 r h = (π /2) [ (31 ) / ( π + 8 ) ] ² + 2 [31 / ( π + 8 )] [62 / (π + 8)]

A(max) = [ 31²/2(π + 8)] = [ 961/ 2( 11.1416 ) ]

A(max) = [ 961/ 22.2832 ] = 43.126 ft²