Optimization in Area

Question

An ostrich farmer wants to enclose a rectangular area and then divide it into five pens with fencing parallel to one side of the rectangle (see the figure below). There are 410 feet of fencing available to complete the job. What is the largest possible total area of the five pens?

Answer 1

possibly you meant one side of the rectangle doesn't need fencing, as it's the side of a barn, house or river? or some other already barricaded side?

IF so, then

410 = L + 6W where area = LW

L = 410-6W

A = LW = (410-6W)W = 410W -6W^2

A'(W) = 410 -12W = 0

W = 410/12 = 205/6 = about 34.17 feet

L = 410-6(410/12) = 410 - 410/2 = 410/2 = 205 feet long

L + 6W = 2205 + 6(410/12) = 205 + 410/2 = 205 + 205 = 410

maximum area = LW = 205 x (205/6) = 205^2/6 = 7,004.1666... square feet

BUT if you meant no side was already in need of fencing, then

2L + 6W = 410

L +3W = 205

L =205 -3W

Area = LW = (205-3W)W = 205W -3W^2 = 410

3W^2 -205W +410 = 0

A'(W) = 6W -205 = 0

W = 205/6 = 34 1/6 feet wide

L = 205 -3(205)/6 = (1/2)205 = 102 1/2 feet long

max area = LW = (205/6)(205/2) = 3052.08333... = about 3,5052 square feet maximum area

which equals half the maximum area if there is no barrier on one side of the rectangular area.