A) A farmer wants to fence in a rectangular plot of land adjacent to the north w

Question

A) A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the barn, and the fencing along the west side side of the plot is shared with a neighbor who will split the cost of that portion of the fence.

If the fencing costs $20 per linear foot to install, and the farmer is not willing to spend more tha $5000, find the dimensions for the plot that would enclose the MOST area.

[Part II]

B) If the Farmer wants to enclose 8000 square feet of land, what dimensions will minimize the cost of the fence?

Explanation / Answer

let the length of the plot be L meters

width of the plot be W meters

Cost of the fence per linear foot = $ 20

==> Total cost of fence = (20L + 20W + (20/2)W)

Since, the fence is not needed on the North side and thefencing is shared by the neighbour on west

==> Cost = 20L + 30W

Now the budget = $5000

==>  20L + 30W = 5000

==> 20L = 5000 - 30W

==> L = (5000 - 30W)/(20)

==> L = 250 - 1.5W

Area of Plot = LW

==> A = (250 - 1.5W)(W)

==> A = 250W - 1.5W2

differentiate with respect to W and equate it to zero

==> dA/dW = (250 - 1.5(2)W2-1) = 0 ; since d/dx xn = n xn-1

==> 250 - 3W = 0

==> W = 250/3

==> L =  250 - 1.5(250/3)

==> L = 250 - 125 = 125

Hence the Dimensions that would enclose maximum area are Length = 125 ft , width = 250/3 = 83.33 ft

b) Cost = 20L + 30W

Area A = LW = 8000

==> W = 8000/L

==> Cost C = 20L + 30 (8000/L)

==> C = 20L + (240000/L)

differentiate with respect to L and equate it to zero

==> dC/dL = 20 - 24000/L2 = 0

==> 240000/L2 = 20

==> L2 = 240000/20

==> L2 = 12000

==> L = 109.54

==> W = 8000/109.54

==> W = 73.03

Hence the dimensions are Length = 109.54 ft and width = 73.03 ft

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.