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.

The area of the rectangle is the product of its dimensions

The area of the rectangle is given as:

Differentiate

Set to 0

Add 2w to both sides

Divide both sides by 2

This means that, the width that gives the maximum area is 20 feet

The perimeter of a rectangle is:

So, we have:

Divide both sides by 2

Subtract 20 from both sides

This means that, the length and the width are equal (i.e. 20 ft)

Hence, the special rectangle is a square

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Width=20 feet

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

Step-by-step explanation:

Area,

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

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.