Paul has 440 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangular that maximize the enclosed area. What is the maximum area?

Answer 1

Perimeter = 440

2L + 2w = 440

L+w = 220

Area = L * w = (220-w)*w = 220w - w^2 = A(w)

-b/(2a) = -220/-2 = 110

Square with side measure 110.

Max area is 12100