# The area of a rectangle whose perimeter is a fixed 80 feet is given by A=40w – w2 , where w is the width of the rectangle. Determine

The area of a rectangle whose perimeter is a fixed 80 feet is given by A=40w - w2 , where w is the width of the rectangle. Determine the width of the rectangle that gives the maximum area. What type of special rectangle is necessary to produce this maximum area? Justify.

## This Post Has 3 Comments

1. whocares1234 says:

Width=20 feet

Since Length=Width=20 feet, the rectangle is a Square.

Step-by-step explanation:

Area, $A=40w - w^2$

To determine the width of the rectangle that gives the maximum area, we take the derivative of A and solve for its critical point.

$A'=40 - 2w\\When A'=0\\40-2w=0\\40=2w\\w=20 feet$

The width of the rectangle that gives the maximum area =20 feet.

Perimeter of a rectangle=2(l+w)

Perimeter of the rectangle=80 feet

2(l+w)=80

2l+2(20)=80

2l=80-40

2l=40

l=20 feet

Since the length and width are equal, the special type of rectangle that produces this maximum area is a Square.

2. Expert says:

something over 10 so ? /10

step-by-step explanation:

3. Expert says: