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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 70.71 ft for the side parallel to the divider and 70.71 ft for the other side.

b)
Dimensions are 56.7 ft for the side parallel to the divider and 88.19 ft for the other side.

c)
Dimensions are 51.59 ft for the side parallel to the divider and 96.92 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 96.92 ft for the side parallel to the divider and 51.59 ft for the other side.

Respuesta :

Answer:

option (d)

Step-by-step explanation:

Data provided in the question:

Let the length parallel to the divider 'L'

and, the whole length of the combined lots perpendicular to the divider 'B'

Fence for the outside boundary costs $7 per foot

Fence for the middle costs $4 per foot

Now,

The area of each lot = 5000 ft²

also,

5000 ft² = [tex]L\times\frac{B}{2}[/tex]

or

B = [tex]\frac{10000}{L}[/tex] ft

Total cost of fencing, C = $4L + 2(B + L) × $7        

substituting the value of B in the above equation:

C = [tex]\$4L + 2(\frac{10000}{L}+L)\times\$7[/tex]

or

C =  [tex]\$18L + \$14\frac{10000}{L}[/tex]

now differentiating the total cost with respect to length,

we get

[tex]C' = 18 - 14\frac{10000}{L^2}[/tex]

for point of minima, put C' = 0

thus,

[tex]18 - 14\frac{10000}{L^2}[/tex] = 0

or

18L² = 140000

or

L = 88.19 ft

Thus,

B =  [tex]\frac{10000}{88.19}[/tex] ft = 113.38 ft

therefore,

Dimensions of the lot is, length parallel to the divider = 88.19 ft

and,

other side = [tex]\frac{B}{2}[/tex] = 56.69 ≈ 56.67 ft

Hence,. The correct answer is option (d)

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