# Optimization problem

• Dec 1st 2011, 06:52 AM
Kjcube
Optimization problem
This problem is giving me some trouble. It's irritating because it seems so easy.

The question is

"The figures are made of rectangles and semicircles.
a.) Find a formula for the area
b.) Find a formula for the perimeter
c.) Find the dimensions x and y that maximize the area given that the perimeter is 100"

The figure is a rectangle with sides x and y with a half-circle connected to one of the ends (diameter of the circle is equal to y).

I got A=xy + 1/2*(1/2y)^2 for the area and
P=2x + y + (1/2)(y)(pi) for the perimeter.

The book gives the answer for c.) as x=100/(4+pi) and y=200/(4+pi)

No matter what I try I can't seem to get their answer. I know I have find the value of x (or y) in terms of the other variable using 100 but I keep getting y=63.661977 and x=31.83098869. I'm stumped.
• Dec 1st 2011, 08:06 AM
corsica
Re: Optimization problem
The area of a (full) circle is $\displaystyle \pi r^2$ (with $\displaystyle r$ being circle radius). I think there's a factor $\displaystyle \pi$ missing in your formula for the area.
• Dec 1st 2011, 08:56 AM
Soroban
Re: Optimization problem
Hello, Kjcube!

Quote:

Code:

              * *  *  *  *  *  *  *           *    |                *         *      |                *       *        | y/2            *                 |                *       *        |                *       *        *                * y       *        |                *                 |                *       *        | y/2            *         *      |                *           *    |                *               * *  *  *  *  *  *  *                 : - - -  x  - - - :
(a) Find a formula for the area.

(b) Find a formula for the perimeter.

(c) Find the dimensions x and y that maximize the area
. . given that the perimeter is 100 inches.

The book gives the answers for (c) as: .$\displaystyle x\:=\:\frac{100}{4+\pi}\,\text{ and }\,y\:=\:\frac{200}{4+\pi}$

(a) Area = (area of rectangle) + (area of semicircle)

. . $\displaystyle A \;=\;xy + \tfrac{1}{2}\pi\left(\tfrac{y}{2}\right)^2 \;=\; xy + \tfrac{\pi}{8}y^2$ .[1]

(b) Perimeter = (perimeter of rectangle) + (perimeter of semicircle)

. . $\displaystyle P \;=\;2xy + \tfrac{1}{2}(\pi y)$

(c) The perimeter is 100 inches: .$\displaystyle 2xy + \tfrac{\pi}{2}y \:=\:100$

. . Solve for $\displaystyle x\!:\;\;x \:=\:\frac{200 - (2+\pi)y}{4}$ .[2]

Substitute into [1]: .$\displaystyle A \;=\;\left(\frac{200-(2+\pi)y}{4}\right)y + \frac{\pi}{8}y^2$

. . which simplifies to: .$\displaystyle A \;=\;50y - \frac{4+\pi}{8}y^2$

Then: .$\displaystyle A' \;=\;50 - \frac{4+\pi}{4}y \;=\;0 \quad\Rightarrow\quad \boxed{y \;=\;\frac{200}{4+\pi}}$

Substitute into [2]: .$\displaystyle x \;=\;\frac{1}{4}\left[200 - (2+\pi)\!\cdot\!\frac{200}{4+\pi}\right]$

. . which simplifies to: .$\displaystyle \boxed{x \;=\;\frac{100}{4+\pi}}$

• Dec 3rd 2017, 04:39 PM
glee12
Re: Optimization problem
How do you end up getting the perimeter of the figure?

"Perimeter = (perimeter of rectangle) + (perimeter of semicircle)

P = 2xy + (1/2)(pi)(y)"

The perimeter of the rectangle itself cannot be '2xy', shouldn't it be '2x + y'?
• Dec 3rd 2017, 05:28 PM
skeeter
Re: Optimization problem
Quote:

Originally Posted by glee12
How do you end up getting the perimeter of the figure?

"Perimeter = (perimeter of rectangle) + (perimeter of semicircle)

P = 2xy + (1/2)(pi)(y)"

The perimeter of the rectangle itself cannot be '2xy', shouldn't it be '2x + y'?

Yes, the sum of three sides of the rectangle that make up part of the figure’s perimeter is 2x+y.