[Note for Mac users] |

Example: Of all rectangles with perimeter 100, which has the greatest area?

- Rewrite the problem with one piece of information per line.
Rectangles

Perimeter 100

Maximize Area

- Draw several pictures illustrating different possibilities.
Some possibilities include a

1 ´ 49 rectangle, a25 ´ 25 rectangle (square), and a30 ´ 20 rectangle.

- Draw a generic picture and introduce variables.
The width and height are varying, so we draw an

*x*´*y*rectangle and let*x*be the width, and*y*the height. Then the area,*A*, is and the perimiter is*A*=*xy* .*P*= 2*x*+ 2*y*= 100

- Explicitly write down "Mazimize ________" or "Minimize ________" and
find a formula for the quantity to be optimized.
Maximize area:

*A*=*xy*

- Use information you know to eliminate all but one variable.
We know

2 , so*x*+ 2*y*= 100 , so*x*+*y*= 50 , hence*y*= 50 -*x* . Now*A*=*x*(50-*x*)*A*is a function of*x*alone.

- Determine the interval where the formula makes sense.
We need

and*x*__>__0 . The latter means*y*__>__050 - , or*x*__>__0 . Thus*x*__<__50*x*must be in the interval [0,50].

- Optimize the function on this interval:
- Find where the derivative is zero.
*A*'(x) = 50 - 2*x*= 0 if and only if .*x*= 25 - Find where the derivative is not defined
The derivative is always defined, in this case.

- Compute the function's value at these points and at the end points of
the interval.
*A*(25) = 25(50-25) = 25 x 25 = 625

*A*(0) = 0(50-0) = 0

*A*(50) = 50(50-50) = 0

- Select the optimum value.
The maximum of 625 occurs at

*x*= 25.

- Compute the quantity requested in the problem.
In this case, we want the dimensions of the rectangle. We have the

*x*value, so we still need*y*. Since , we have*y*= 50 -*x* . So the rectangle of perimeter 100 with the greatest area is a*y*= 50 - 25 = 2525 ´ 25 rectangle.

- Find where the derivative is zero.

Comments to: dpvc@union.edu

Created: Oct 20 1997 --- Last modified: Oct 27, 1998 3:49:33 PM