Minimum cost.From a tract of land,a developer plans to fence a rectangular region and then divide it into two identical rectangular lots by putting a fence down the middle.Suppose that the fence for the outside boundary costs $7 per foot and the fence for the middle costs $4 per foot.If each lot contains 5000 square feet,find the dimensions of each lot that yield the minimum cost for the fence.
A) Dimensions are 51.59 ft for the side parallel to the divider and 96.92 ft for the other side.
B) Dimensions are 96.92 ft for the side parallel to the divider and 51.59 ft for the other side.
C) Dimensions are 70.71 ft for the side parallel to the divider and 70.71 ft for the other side.
D) Dimensions are 88.19 ft for the side parallel to the divider and 56.7 ft for the other side.
E) Dimensions are 56.7 ft for the side parallel to the divider and 88.19 ft for the other side.

Question

Minimum cost.From a tract of land,a developer plans to fence a rectangular region and then divide it into two identical rectangular lots by putting a fence down the middle.Suppose that the fence for the outside boundary costs $7 per foot and the fence for the middle costs $4 per foot.If each lot contains 5000 square feet,find the dimensions of each lot that yield the minimum cost for the fence. ​
A) Dimensions are 51.59 ft for the side parallel to the divider and 96.92 ft for the other side.
B) Dimensions are 96.92 ft for the side parallel to the divider and 51.59 ft for the other side.
C) Dimensions are 70.71 ft for the side parallel to the divider and 70.71 ft for the other side.
D) Dimensions are 88.19 ft for the side parallel to the divider and 56.7 ft for the other side.
E) Dimensions are 56.7 ft for the side parallel to the divider and 88.19 ft for the other side.

Quizplus · Accepted Answer

Let x be the width of the lot parallel to the fence down the middle, and let y be the other width. Then the total cost of the fence is:

C = 2(7x) + 2(7y) + 4x = 14x + 14y + 4x = 18x + 14y

The total area of the two lots is:

A = 2(xy) = 2(5000) = 10000

Solving for y in terms of x gives:

y = 10000/2x = 5000/x

Substituting this into the cost equation gives:

C(x) = 18x + 14(5000/x)

To find the minimum cost, take the derivative and set it equal to zero:

C'(x) = 18 - 700000/x^2 = 0

Solving for x gives x = 56.7 ft, and substituting back to find y gives y = 88.19 ft. Therefore, the dimensions of each lot are 56.7 ft by 88.19 ft, and the minimum cost is:

C(min) = 18(56.7) + 14(88.19) = $2026.46

Minimum Cost