Need help with Calculus

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 28 feet?

Answer 1

A = A_rectangle + A_semicircle

A = w•h + 1/2(π)(w/2)²

A = wh + πw²/8

The perimeter is w + 2h + 1/2(2π(w/2)) = w + 2h + (π/2)w and, since the perimeter is 28, we can say:

w + 2h + (π/2)w = 28

2h + w(1 + π/2) = 28

2h + w(2 + π)/2 = 28

2h = 28 - w(2 + π)/2

h = 14 - w(2 + π)/4 then, plugging this into the area equation we get:

A = wh + πw²/8

A(w) = w(14 - w(2 + π)/4) + πw²/8

A(w) = 14w - w²(2 + π)/4) + πw²/8

A(w) = 14w - w²[(2 + π)/4) + π/8]

A(w) = 14w - w²[1/2 + 2π/8 + π/8]

A(w) = 14w - w²[1/2 + 3π/8]

Taking the derivative:

A'(w) = 14 - 2w(1/2 + 3π/8)

A'(w) = 14 - w(1 + 3π/4)

Setting the derivative equal to zero:

14 - w(1 + 3π/4) = 0

14 = w(1 + 3π/4)

w = 14/(1 + 3π/4)

w ≈ 4.171 ft

Plugging that into our area function:

A(4.171) = 14(4.171) - (4.171)²[1/2 + 3π/8] = 29.2 sq ft