Optimization Problem?

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 42 feet?

Answer 1

Let the length of the window be " p " and the radius of the semicircle be " r " (the width of the rectangle is 2r)

the perimeter, P=2 sides (length) + 1 side width +semicircumference
so, P=2p+2r+πr
2p+(π+2)r=42 (given)..........(1)

The area, A=A(semicircle)+A(rectangle)
=(πr^2)/2+2pr...........(2)

From the (1), we have 2p=42-(π+2)r

Inserting this into (2) we have A=(πr^2)/2+(42r-(π+2)r^2)
Rearranging: A=42r-(π/2 +2)r^2

To optimize the area, we need to find r for which dA/dr=0:
dA/dr =42-(π+4)r=0

Hence, r=42/(π+4) inserting this value into A gives
A= 882/(π+4) ft^2

btw, thanks for refreshing my memory...I forgot what those windows were called...

Answer 2

First, we need the perimeter of the window. It is equal to

P = (perimeter of 3 sides of a rectangle) + (half the circumference of a circle).

Let x be the width and y be the height. By extension, if x is the width of the rectangle, then it is also the diameter of the semicircle.

P = x + y + y + (pi)x

P = (pi + 1)x + 2y

But P = 42,

42 = (pi + 1)x + 2y

The area of this window is going to be

A = (area of rectangle) + (area of semicircle)

The area of the rectangle is just xy.
The area of the semicircle is half the area of a circle, and the radius is x/2 (since the diameter is x). Thus, the area of a full circle with radius (x/2) is equal to pi(x/2)^2, or pi(x^2/4).

We have to cut this in half because it is a semicircle, so the area of the semicircle is (1/2)pi(x^2/4), or
(1/8)pi(x^2).

Combining those two,

A = xy + (1/8)pi(x^2)

Solve for y in the previous perimeter equation:

42 = (pi + 1)x + 2y

42 - (pi + 1)x = 2y
(1/2) (42 - (pi + 1)x) = y

A = x[(1/2) (42 - (pi + 1)x)] + (1/8)pi(x^2)

Let's clean that up.

A = (x/2) (42 - (pi + 1)x) + (1/8)pi(x^2)
A = 24x - (pi + 1)(x^2)(1/2) + (1/8)pi(x^2)

A = 24x + (1/8)pi(x^2) - (1/2)(pi + 1)(x^2)

A = 24x + x^2 ( (1/8)pi - (1/2)(pi + 1) )

A = 24x + x^2 ( (1/8)pi - (1/2)pi - (1/2) )

A = 24x + x^2 ( (pi/8) - (4pi/8) - (4/8) )

A = 24x + x^2 ( (-3pi - 4)/8 )

A = 24x + (1/8)(-3pi - 4)x^2

This will be our area function.

A(x) = 24x + (1/8)(-3pi - 4)x^2

Take the derivative.

A'(x) = 24 + 2x (1/8)(-3pi - 4)
A'(x) = 24 + x (1/4)(-3pi - 4)

Make A'(x) = 0.

0 = 24 + x (1/4)(-3pi - 4)

Solve for x.

-24 = x (1/4)(-3pi - 4)

-96 = x (-3pi - 4)

96 = x(3pi + 4)

x = 96/(3pi + 4)

To find the maximum area, plug in this value for A(x).

A(x) = 24x + (1/8)(-3pi - 4)x^2
A(x) = (1/8)x (192 + (-3pi - 4)x )
A(x) = (1/8)x (192 - (3pi + 4)x )

A( 96/(3pi + 4) ) = (1/8)[96/(3pi + 4)] (192 - 96)

= (1/8) (96/(3pi + 4) ) (96)
= (9216)/8 (1/(3pi + 4))
= 1152/(3pi + 4)

Answer 3

Let r be the radius of the semicircle, atop a rectangle of width w = 2r and height h.

You want to maximize the total area (1/2) pi r^2 + 2 h r subject to the contraint pi r + 2h + 2r = 42 feet (and that r, h, w are nonnegative).

With Lagrange multiplier y we let B = (1/2) pi r^2 + 2 h r + y (pi r + 2 h + 2 r - 42) and set:
dB/dr = 0 = pi r + 2h + y (pi + 2)
dB/dh = 0 = 2 r + 2 y
dB/dy = 0 = pi r + 2 h + 2 r - 42

You get y = -r from dB/dh = 0, and plugging that into dB/dr = 0 gives pi r + 2 h - r (pi + 2) = 0; or 2 h - 2 r = 0, so h = r. Finally the perimeter constraint yields pi r + 2 r + 2 r = 42, so r = 42 / (4 + pi) or approx. 5.881, and the maximum area is (1/2) pi r^2 + 2 h r = (1/2) pi r^2 + 2 r^2 = (pi/2 + 2) r^2 = (1/2) (pi + 4) (42 / (4 + pi))^2 = (1/2) 42^2 / (pi + 4) or approx. 123.5 square feet.

Dan