Norman window has shape of a semicircle atop a rectangle so that the diameter of the semicircle is = to width of the rectangle. What is area of the largest Norman window with perimeter of 45 feet?

Answer 1

Maximizing area for a given perimeter P (a general solution)

where r = the radius of the semicircle

Let x = 2r + 2L be the part of the perimeter that comes from the 3 sides of the rectangle.

P = πr + x

L = ½(x – 2r)

If you maximize the rectangular area (A₂), you maximize the whole area A = A₁ + A₂

Maximizing the rectangular area A₂

A₂ = 2rL = r(x – 2r) = rx – 2r²

dA₂/dt = x – 4r = 0 means that x = 4r

L = ½(x – 2r) = ½(4r – 2r) = ½(2r)

So area is maximized when L = r = ½W

P = πr + x = πr + 4r = (π + 4)r

which means that A₂ is maximized when

L = r = P/(π + 4)

W = 2r = 2P/(π + 4)

Maximum Total Area

A = A₁ + A₂ = ½πr² + 2rL = ½πr² + 2r² = ½(π + 4)r² = ½P²/(π + 4)



In the current problem P = 45 ft, so

L = r = 45/(π + 4) = 6.3 ft

W = 2(45)/(π + 4) = 90/(π + 4) = 12.6 ft

A = ½P²/(π + 4) = ½(45)²/(π + 4) = 141.8 ft²