David has available 3600 yards of fencing and wishes to encl

David has available 3600 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area?

Solution

Total length of fencing = 3600 yards

It is used to enclose a rectangular region

Let L be length of region , W be width of region

==> Perimeter of region is 2(L + W)

==> 2(L + W) = 3600

==> L + W = 3600/2

==> L + W = 1800

==> L = 1800 - W

Area of region = LW

==> A(W) = (1800 - W)W

==> A(W) = -W² + 1800W

Area is maximum ==> A \'(W) = 0

==> -2W^2-1 + 1800(1) = 0         since d/dx xⁿ = nx^n-1

==> -2W + 1800 = 0

==> W = 1800/2

==> W = 900

==> L = 1800 - 900 = 900

Hence Maximum Area A = LW = (900)(900)

==> A_max = 810000 square yards