Pre-Calculus Test Review problem

Question

Norman Window: A Norman window is a window that has the shape of a semi-circle on a rectangle. A Norman window is going to be built with the perimeter of 30 ft. The diameter of the circle is equal to the width of the rectangle. Determine the dimensions of the Norman window that allows the maximum amount of light to pass through.

Answer 1

First draw the diagram of a rectangle with a semicircle on top. Call the width of the rectangle w and the height h. So the diameter of the circle is w and its radius is w/2.

Write a formula for the perimeter using half the circumference of the circle.

P: 2h + w + (.5(2πw/2) = 30

Simplify 2h + w + (π / 2) w = 30

Multiply by 2 to clear the fraction. 4h + 2 w + π w = 60

Solve for 4h 4h = 60 - 2w - π w

Divide both sides by 4 h = (60 - 2w - π w)

4

Next write an equation for the Area using the formula for half the area of a circle.

A = wh + .5 ( π (w/2)²) Now substitute the above solution for h into this equation.

A = w ( (60 - 2w - π w)) + .5 ( π (w/2)²) Simplify by separating each term.

4

A = 15 w - 2 w² - π w² + π w ² The last 2 terms now collect -1/4 + 1/8 = - 1/8

4 4 8

A = 15 w - 1/2 w² - 1/8 π w²

I put this quadratic equation into DESMOS to draw the parabola.

The vertex lies at (8.4, 63). This means that the width is 8.4 and the max area is 63.

I checked it by putting 8.4 into the above equation for h. I got h = 4.2

I calculated the area of the rectangle and the semicircle and got 62.9888.

The discrepancy is rounding error.

Answer 2

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 = 30 means that 2L = 30 – (1 + ½π)W and

L = 15 – (½ + ¼π)W = 15 – (π + 2)W/4

This gives

Area as a Function of W

A = (15 – (½ + ¼π)W)W + ⅛πW² = 15W + (⅛π – ½ – ¼π)W² = 15W – (½ + ⅛π)W²

Value of W Giving Maximum Area

Since – (½ + ⅛π) < 0, this function describes a parabola opening downward and

its maximum occurs at its vertex point at W = –b/2a where b = 15 and a = –(½ + ⅛π).

The vertex point occurs at W = 15/[2(½ + ⅛π)] = 60/(π + 4) ≈ 8.40 ft

Solving for L

L = 15 – (π + 2)W/4 = 15 – 60(π + 2)/4/(π + 4) = 15 – 15(π + 2)/(π + 4) = 15[1 – (π + 2)/(π + 4)] =

15[(π + 4 – π – 2)/(π + 4)] = 15[2/(π + 4)] = 30/(π + 4) ≈ 4.20 ft

Note that L is exactly half of W

Answer

The dimensions of the Norman window that allows the maximum amount of light to pass through

W = 60/(π + 4) ≈ 8.4 ft

L = 30/(π + 4) ≈ 4.2 ft

Check

P = 2L + W + ½πW = 30

P = (2×30)/(π + 4) + 60/(π + 4) + (½π×60)/(π + 4) = (60 + 60 + 30π)/(π + 4) = (120 + 30π)/(π + 4)

= 30(4 + π)/(π + 4) = 30