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.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
I found the base length that is equal to 60/(4+pi) ft but I can't find the total height. Thank you!!

Answer 1

One of the popular questions in Calculus texts.

Let the radius of the circle be x
then the width of the rectangle = 2x
let the height of the rectangle be y

2x + 2y + halfcircle = 30
2x + 2y + (1/2)(2πx) = 30
2x + 2y + πx = 30
2y = 30 - 2x - πx
y = 15 - x - πx/2

area of window = A
= 2xy + (1/2)πx^2
= 2x(15 - x - (1/2)πx) + (1/2)πx^2
= 30x - 2x^2 - πx^2 + (1/2)πx^2

dA/dx = 30 - 4x - 2πx + πx
= 0 for a max of A
30 = 4x+πx
30 = x(4+π)
x = 30/(4+π) = appr 4.2

base = 2x = 60/(4+π) ---> good for you, you had that

now y = 15 - 4.2 - π(4.2)/2 = appr 4.2 , ahhh, the same

So the rectangle is
60/(4+π) by 30/(4+π)
or
appr 8.4 at the base and 4.2 high