Answer to Question #270244 in Calculus for Mie

Question

A Norman window consists of rectangle surmounted by a semicircle. If the

perimeter of a Norman window is 32 ft, what should be the radius of the

semicircle and the height of the rectangle such that the window will admit most

light?

Answer 1

Let x denote half the width of the rectangle (so x is the radius of the semicircle), and let h denote the height of the rectangle.

perimeter of the window:

$2y + 2x +2πx/2 = 2y + (2 + π)x=32$

then:

$y=16-\frac{x(\pi +2)}{2}$

 area of the window:

$A=2xy+\pi x^2/2$

$A(x)=2x(16-\frac{x(\pi +2)}{2})+\pi x^2/2=32x-\frac{x^2(\pi +4)}{2}$

$A'(x)=32-(4+\pi)x=0$

radius of the semicircle such that the window will admit most light:

$x=\frac{32}{4+\pi}=4.48$ ft

height of the rectangle such that the window will admit most light:

$y=16-\frac{4.48(\pi +2)}{2}=4.48$ ft