If the perimeter of the window is 24 ft, find the value of x so that the greatest possible amount of light is admitted.

Question

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 24 ft, find the value of x so that the greatest possible amount of light is admitted.

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)

In the current problem P = 24 ft, so

W = 2(24)/(π + 4) = 48/(π + 4) = 6.72 ft

Answer 2

First, let's draw a picture of this window, which is essentially a rectangle with h height with a semi circle top that has a diameter of w width. We want to find a window with the largest area possible.

Now let's determine the area, which is just the area of the rectangle part plus the area of half the circle. This is

A = h * w + ( π r ²)/2

Remember our radius is just half the width, so

A = h * w + ( π (w/2) ²)/2

The area, or amount of light being let in, is what we want to maximize, but first let's make our area function A into a single variable function.

To do that we find our constraint, which tells us the perimeter of the whole window must be 2h + π r + w = 24 , where π r is the perimeter of a semi circle Remember that the radius is w/2, so

24 = 2h + π w/2 + w

Now we put this into the area function A by isolating h

h = [24 - π w/2 - w]/2

A = [24 - π w/2 - w]/2 * w + ( π (w/2) ²)/2

Expand it out and get

A = 1/2 (24 w - w² - (π w²)/ 2) + π w²/8

So to find the max w would mean to find the derivative of A

A' = 1/2 (24 - 2w - π w) + π w/4

A' = 12 - w - π w/4

And set this equal to 0 to find your maximum w.

12 - w - π w/4 = 0

Solve for w to get

w = 48/ (4 + π ) = 6.72

So the width of the window that will allow the most light is w = 6.72 ft. The problem labeled the width x, so x = 6.72 ft.