Afarmer 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

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

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.

72 is the answer to this problem