Question

A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. what is the largest possible total area of the four pens

Answer 1

Answer:

22,562.5 ft²

Step-by-step explanation:

The largest possible area will be obtained when half the fencing is used for the long side of the 4 pens, so the dimension in that direction is (950 ft)/(2·2) = 237.5 ft.

The other half of the fencing will be used for the 2 ends and 3 partitions, each of which will be (950 ft)/(2·5) = 95 ft.

Then the overall area of the 4 pens is ...

... (237.5 ft)(95 ft) = 22,562.5 ft²

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General Solution

Suppose L is the length of fence available, and x is the length of the long side of the enclosed area. For n pens, the enclosed area will be ...

... A = x(L-2x)/(n+1)

For constant values of A, L, n, this describes a downward-opening parabola with zeros at x=0 and x=L/2. The vertex of the parabola (point of maximum area) will be halfway between these zeros, at x = (0 + L/2)/2 = L/4. That is, half the available fence is used in each of the orthogonal directions.

Note that adding partitions in the other direction replaces the 2 in the equation with (m+1), where m is the number of pens between the sides of length x. That is, if there are 4 pens in one direction by 3 pens in the other direction, the area will be

... A = x(L -(3+1)x)/(4+1)

and, once again, we find that half the fence is used in each of the orthogonal directions when we maximize the overall area.