Two equal rectangular lots are enclosed by fencing the perim

Two equal rectangular lots are enclosed by fencing the perimeter of a rectangular lot and then putting a fence across its middle. If each lot is to contain 300 square feet, what is the minimum amount of fence needed to enclose the lots (include the fence across the middle)? ft A rectangular area is to be enclosed and divided into thirds. The family has $520 to spend for the fencing material. The outside fence costs $10 per running foot installed, and the dividers cost $20 per running foot installed. What are the dimensions that will maximize the area enclosed? (The answer contains a fraction.) a side parallel to the dividers ft a side perpendicular to the dividers ft

Solution

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Let the dimension of the each rectangular lot be x ft ( length) by y ft (breadth). Then xy = 300 or, y = 300/x….(1) The length of the large rectangular plot comprising of both the plots is x and its breadth is 2y. Therefore, the length of the fencing required to enclose this large plot is 2 ( x + 2y) = 2x + 4y . Now, the length of the fence across the middle of the large plot is x. Therefore, the length of fencing required is x + 2x + 4y = 3x + 4y = 3x + 4*300/x =    3x + 1200/x = P (say). Now, P will be minimum when dP/dx = 0 i.e. when 3 – 1200/x² = 0 i.e. when 1200/x² = 3 i.e. when 3x² = 1200 i.e. when x² = 400, i.e. when x = 20. Then P = 3x + 1200/x = 3*20 + 1200/(20) = 60 + 60 = 120 ft. Thus, the minimum amount of fence required is 120 ft.

2. Let the dimension of the rectangular lot be x ft ( length) by y ft (breadth). The perimeter of the plot is 2(x + y) ft. The cost of fencing the perimeter is 10* 2(x + y) = 20(x +y) . Since the plot is to be divided into thirds, there will be a requirement of 2 more fencings by way of dividers. The length of each divider being x, the total length of the dividers is 2x and its cost is 20*2x = 40x Therefore, 20 (x + y) + 40x = 520 or, 60x + 20 y = 520 or 3x + y = 26 or, y = 26 – 3x …(1)

The area A, of the rectangular plot (original large plot before dividing) is xy = x (26 -3x) = 26x – 3x² . Now, A is maximum when dA/dx = 0 i.e. when 26 - 6x = 0 i.e. when 6x = 26 or, when x = 26/6 = 13/3 ft. Then the area of the plot is 26x – 3x² = 26 (13/3) – 3(13/3)² = 338/3 – 169/3 = 169/3 sq.ft. The side parallel to the dividers is x = 13/3 ft. The side perpendicular to the dividers is 26 – 3x = 26 - 3*13/3 = 26 -13 = 13 ft.

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