A rectangular fence is to be built with 325 yds of fencing. The total A enclosed within the rectangle can be expressed by the function A(l)=-2l^2+325l where l is the length of the rectangle (in yds)

Question

Find the rectangle length that yields the maximum area. What is the maximum area? Explain.

Answer 1

A(L) = -2L² + 325L

The graph of this function is a downward facing parabola since the coefficient of the squared term is negative. Solving for L, if A(L)=0 then 0=-2L²+325 gives two L-intercepts at L=0 and L=162.5 yards. The vertex, which is the maximum value for L for this parabola, is halfway between these values, so L_max = 81.25 yards.

This can also be derived from the derivative A^'(L)=-4L+325. The derivative = 0 for the maximum value of A. If 0=-4L+325, then L_max=325/(4) = 81.25 yards.

Another way to look at this problem is to realize the rectangle with the largest area for a fixed perimeter is a square, so L_max is the length of the side of a square and the perimeter = 325 yards. 4L = 325, so L_max = 81.25 yards. The area enclosed by this rectangle (when it is a square) = (81.25 yards)² = 6,601.5625 square yards.

Notice that this is 1/2 the value of the (incorrect) formula for area A(L) = -2L² + 325L evaluated for L=81.25 yards. This is because the correct formula is derived from 2L + 2W = 325.

2W = -2L + 325

W = (-2L + 325)/2

A(L) = L x W = L x (-2L +325)/2

A(L) = (-2L² +325L)/2