Find the largest possible window

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 20 feet?

Answer 1

Area

Area of the rectangle: A_rect = L × W

Area of the semi-circle: A_s-c = ½πr² = ½π(½W)² = ⅛πW²

Area of the Norman window: A = A_rect + A_s-c

A = L × W + ½π(½W)² = L × W + ⅛πW²

Using Perimeter to find L as a Function of W

2L + W + ½πW = 2L + (1 + ½π)W = 20 means that 2L = 20 – (1 + ½π)W and

L = 10 – (½ + ¼π)W

This gives

Area as a Function of W

A(W) = (10 – (½ + ¼π)W)W + ⅛πW² = 10W + (⅛π – ½ – ¼π)W² = 10W – [(π+4)/8]W²

Value of W Giving Maximum Area

Since –(π+4)/8 < 0, this function describes a parabola opening downward and

its maximum occurs at its vertex point at W = –b/2a where b = 10 and a = –(π+4)/8.

The vertex point occurs at W = 10/[2(π+4)/8)] = 40/(π + 4)

Setting dA/dt = 10 – [(π+4)/4]W = 0 also gives the same value W = 40/(π + 4) ≈ 5.6 ft

Maximum Area

Evaluate A[40/(π + 4)] to find A ≈ 28 ft²

Answer 2

Let D = diamter of semicircle top, H= height of rectangular portion and P=perimeter=20=πD/2+2H+D

Let A = Area of window = πD²/4+DH

Solve 20=πD/2+2H+D for H

Substitute result into Area

Take derivative of result to find dA/dD=πD-2D+10 and set = 0 to find D=10/(π-2)

Substitute into A to find A = 15

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