Question

Jacob 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 Jacob has 400 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).

A(w) =

b) What width
w would maximize the area?
w = _____ ft
c) What is the maximum area?
A = _______square feet

Answer 1

Jacob 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.

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).

one side of the length is not counted for perimeter because one side of length will be against the barn.

Perimeter = 400 ft

Perimeter of rectangle = L + W + W

400 = L + 2W

L = 400 - 2W

Area = L * W

Replace L by  400 - 2W

A(W) =  (400 - 2W) * W

$A(W) = -2W^2 + 400W$

Now we find out x coordinate of vertex to find the width that maximize the area

$W = \frac{-b}{2a}$

a= -2 and b = 400

$W = \frac{-(400)}{2(-2)}=100$

The width  w would maximize the area is w = 100ft

To find maximum area we plug in 100 for W in A(W)

$A(W) = -2W^2 + 400W$

$A(W) = -2(100)^2 + 400(100)= 20000$

the maximum area is 20,000 square feet