Question

Mr. Harrell has 64 feet of fencing that he will use to build a rectangular enclosement for chickens and he will use his barn as one of of the sides of the rectangle as seen in the image. Write a function A (x) relating the area of the enclosement A and the width of the rectangle x. Then, find the width that maximizes the area. Then state the maximum area.

Answer 1

• A = x(64 -x)/2
• x = 32 ft
• 512 ft²

Explanation:

Given 64 feet of fencing that can be used for three sides of a rectangular enclosure next to a barn wall, you want a function for area in terms of enclosure width, the width that maximizes area, and the maximum area.

(a) Function

If we define the width (x) as the dimension parallel to the barn wall, then the fence available for the "height" (out from the barn wall) is (64-x) in length. Half that is used on either end of the enclosure, so the area is ...

A = WH

A(x) = x(64 -x)/2

(b) Best width

We notice this function describes a parabola that opens downward, with zeros at x=0 and x=64. The vertex (maximum) is on the line of symmetry, halfway between the zeros:

x = (0 +64)/2 = 32

The width that maximizes area is x = 32 ft.

(c) Maximum area

Using x=32 in the area formula, we find the maximum area to be ...

A(32) = 32(64 -32)/2 = 1024/2 = 512 . . . . square feet

The maximum area is 512 square feet.

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Additional comment

Perhaps you can see that the best width is half the length of the available fence. The other half of the fence is used for the ends of the enclosure. This is the generic solution to this kind of problem.