Optimization: maximizing area?

Question

An open rectangle with a semicircle on the top has a total perimeter of 288 inches. Find the radius of the semicircle that will maximize the area of the window.

Answer 1

Let us denote L to be the length of the window and W be the width. Then, our perimeter will be denoted as follows:

P = (three sides of the rectangle) plus (semicircle)

Getting the three sides of the rectangle is simple:
It's just L + W + L.
Getting the semicircle will just be getting half the circumference of the circle. If you drew a diagram, you'll notice W represents the diameter of the circle. Since circumference equals pi times diameter, and the diameter is W, then half the circumference would be W*pi/2

So

P = L + W + L + W*(pi/2)
P = 2L + (1+pi/2)W

Since we ultimately want the radius, and since we can see that W is equal to the diameter, it turns out W=2r, so we change the equation around

P = 2L + (1+pi/2)(2r)
P = 2L + (2 + pi)r
But P = 288, so

2L + (2 + pi)r = 288

Now, we want to maximize the area of the window, so we need a formula for the area.

A = (area of the rectangle) plus (area of the semicircle)
A = L(2r) + (pi*r^2)/2
A = 2Lr + (1/2)pi*r^2

Now, note that for
2L + (2 + pi)r = 288
We can solve for one variable; let's solve for L.

2L = 288 - (2+pi)r
L = 144 - [(2+pi)/2] r

So now, we plug in L for our area function.

A = 2 [144 - [(2+pi)/2] r] + (1/2)pi*r^2

And this is now our function, which we'll from now on denote as
A(r), since it is now a function of ONE variable.

A(r) = 288 - (2+pi)r + (1/2)pi*r^2

In order to maximize, we must take the derivative and make it zero.

A'(r) = - (2+pi) + pi*r
0 = -2 - pi + pi*r
2 + pi = pi*r
r = (2+pi)/pi

Therefore, r = (2+pi)/pi is the radius of the semicircle that will maximize the area of the window.

Answer 2

The perimeter = bottom of rect. + semicircle + sides = 2r + pi*r + 2 * h = 288.

Solve for h. h = (288 - 2r - pi*r) / 2

The area is 2rh + 0.5*pi*r^2. Put this in terms of r by subbing in for h.

A(r) = 2r((288 - 2r - pi*r) / 2) + 0.5*pi*r^2 = 288r - (2 + 1.5 pi)r^2

The max happens where the derivative = 0.

A'(r) = 288 - (4 + 3pi)r = 0.

Solve. (I hope I didn't have any typos, but this is the approach.)

Answer 3

If 2L = length of rectangle and 2w = its width, then the radius of the circle part if it's attached to the width side is w.

Perimeter = 2L + 2w + 2L + 2w(pi)/2 = 288

4L + 2w + w(pi) = 288

4L = 288 - 2w - w(pi)

L = 72 - 1/2 w - w(pi)/4

Area = 2L(2w) + w^2 pi

Substitute in for L, then find the derivative and slove ot for 0