Question

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the perimeter of the window is 32 ft, find the value of x so that the greatest possible amount of light is admitted.

Answer 1

Let x ≡ width of the rectangle, y ≡ height of the rectangle
 
A = xy + π/8 x²  area: rectangle + semicircle
x + 2y + π/2 x = 32 perimeter: rectangle + semicircle
 
A = x(16 - x/2 - π/4 x) + π/8 x²
dA/dx = 16 - x/4 (4 + π) 16 = x/4 (4 + π) set dA/dx = 0 to find stationary points
x = 64/(4 + π) is a relative max since d²A/dx² < 0
y = 32/(4 + π)
 
Answer: 64/(4 + π) or approx. 8.96 ft (to 2 d.p.)

Answer 2

The value of x so that the greatest possible amount of light is admitted is 4.48 ft.

Let x represent the radius of the semicircle, hence the width of the rectangle = 2x. Also, the length of the rectangle is y

The perimeter of the window = πx + 2y + 2x = x(π + 2) + 2y

32 = x(π + 2) + 2y

$y=\frac{32-x(\pi +2)}{2}$

The area of the window (A) = 0.5πx² + 2yx$A=\frac{1}{2}\pi x^2 +2x(\frac{32-x(\pi + 2)}{2} )\\\\A=\frac{1}{2}\pi x^2 +32x-\pi x^2-2x^2\\\\A=32x-x^2(2+\frac{\pi}{2} )\\\\The \ maximum\ value\ of\ x\ is\ at\ dA/dx=0\\\\A'=32-2x(2+\frac{\pi}{2} )\\\\2x(2+\frac{\pi}{2} )=32\\\\x(2+\frac{\pi}{2} )=16\\\\x=4.48\ ft$

Hence the value of x so that the greatest possible amount of light is admitted is 4.48 ft.

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