Farmer Brown has 750 yards of fencing with which to build a rectangular corral.

Question

Farmer Brown has 750 yards of fencing with which to build a rectangular corral. He wants to divide it evenly into three pens, so he adds in two divider fences, as shown below. Answer the following:

a) Write the area of the corral as function of x 

b) Determine the maximum area enclosed by the corral. (Decimal approximations are marked incorrect.)

Answer 1

Let l = the length of the corral and let w = the length of the 4 equal pieces that divide the corral into 4 equal sections. Then we know from the amount of fencing available that

2 x l + 4 x w = 750 We also know that the Area is given by A = l x w

So we have from the first equation: 4 x w =775 - 2 x l or w = 193.75 - 1/2 x l Substituting into the Area

equation we have:

A = 193.75 x l - 1/2 x l² To find the maximum area differentiate and set equal to zero

da/dl = 193.75 - l = 0 So l = 193.75 and

w = 193.75 - 1/2 x l = 96.875 all units in yards which can be converted to feet and inches.

Answer 2

Let l = the length of the corral and let w = the length of the 4 equal pieces that divide the corral into 4 equal sections. Then we know from the amount of fencing available that

2 x l + 4 x w = 750 We also know that the Area is given by A = l x w

So we have from the first equation: 4 x w =775 - 2 x l or w = 193.75 - 1/2 x l Substituting into the Area

equation we have:

A = 193.75 x l - 1/2 x l² To find the maximum area differentiate and set equal to zero

da/dl = 193.75 - l = 0 So l = 193.75 and

w = 193.75 - 1/2 x l = 96.875 all units in yards which can be converted to feet and inches.

Answer 3

We start with 2 lengths and 2 widths for the area, then add 2 lengths to separate the pens into 3.

Thus, the amount of fencing required is 4l + 2w, and the area is lw

As we have 750 yards of fencing, 4l + 2w = 750.

w = (750-4l)/2 = 375 - 2l

Area is lw = l(375-2l) = 375l - 2l^2

If we are in calculus, we take derivatives to maximize; if we are in Algebra, we just complete the square.

A = -2(l^2 - 375l/2) = -2(l^2 - 375l/2 + (375/4)^2) + 2(375/4)^2

This is maximized at l = 375/4, and the maximum area is 2(375/4)^2 = 375^2/8 = 17578 1/8

I don't have a picture, so I can't be sure what x refers to; l is 375/4 and w = 375/2.

Answer 4

Draw and label a diagram!

x represents the width of the corral and the two separators

750 - 4x represents the length of the corral

A = x(750 - 4x)

A = 750x - 4x²

Maximum area occurs at x = 750 / 8 or 93.75