Question

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.

Answer 1

Answer:

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.

Answer 2

The area of the rectangle is the product of its dimensions

• The width that gives the maximum area is 20 feet
• The special rectangle is a square

The area of the rectangle is given as:

$\mathbf{A = 40w - w^2}$

Differentiate

$\mathbf{A' = 40 - 2w}$

Set to 0

$\mathbf{40 - 2w = 0}$

Add 2w to both sides

$\mathbf{ 2w = 40}$

Divide both sides by 2

$\mathbf{ w = 20}$

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

The perimeter of a rectangle is:

$\mathbf{P = 2(l + w)}$

So, we have:

$\mathbf{80 = 2(l + 20)}$

Divide both sides by 2

$\mathbf{40 =l + 20}$

Subtract 20 from both sides

$\mathbf{l = 20}$

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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