Derek wants to build a rectangular enclosure for his animals.

Question

Derek wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Derek has 850 feet of fencing, you can find the dimensions that maximize the area of the enclosure.

a) Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn). Write an function for the area A of the enclosure in terms of w. (HINT first write two equations with w and l and A. Solve for l in one equation and substitute for l in the other).

b) What width w would maximize the area?

c) What is the maximum area?

Answer 1

First of all the maximum area is always a square. You can get the answer first to check your work later.

850/4=212.5 feet

One side is not needed so add that length to the non-parallel side.

212.5+212.5=425 feet is the length and 212.5 feet is the width(s)

A=425*212.5

A=90,312.5 square feet

Doing the algebra...

P=l+2w

850=l+2w

850-2w=l

A=lw

A=(850-2w)w

A=850w-2w²

This is a parabola that opens downward so there is a maximum point.

The vertex of the parabola is (h,k) where h is the "maximizing number" and k is the maximum area.

Use the fact that h=-b/2a

h=-850/(2*[-2])

h=-850/(-4)

h=212.5 would be the length of all four sides if it were not for the barn

Therefore you have an extra 212.5 feet

Add the 212.5 feet to the opposite side(length) to get 425 feet.

You have a rectangle that is 212.5 feet by 425 feet by 212.5 feet by "the barn".

The width is 212.5 feet which maximizes the area.

A=lw

A=425*212.5

A=90,312.5 square feet is the maximum area.

Note that if you substituted 212.5 for w in the equation A=850w-2w² you would get 90,312.5. So 212.5 is the maximizing number, h, and the maximum area is k, which is 90,312.5 square feet.