Optimization Solutions

Question

A farmer is setting up separate pens for livestock by fencing off a rectangular area and running fence to create four parallel pens. If the farmer has a total of 200 feet of fence, what is the largest area that can be contained?

Answer 1

20 ft by 50 ft

max Area = 1,000 ft^2

2 sides in one direction, call it the Length

5 sides in a perpendicular direction, call them the Width

200 total feet of fencing

200 = 2L + 5W

A = LW = (200-5W)/2](W) = 100W - 5W^2/2

take the derivative of A with respect to W and set = 0

dA/dW=A'(W) = 100 -5W = 0

W = 100/5 = 20 feet Wide

L = (200-5W)/2 = (200-5(20)/2 = 100/2= 50 feet

max Area = 20x50 = 1,000 ft^2

each of the 4 pens has 4 sides, 2 sides =50 ft and 2 sides =20 ft

by putting the pens adjoining each other they 2 pens share 1 side

and the middle 2 pens share 2 sides, so you end up with 5 sides

in one direction and 2 sides in the perpendicular direction.

that leaves 200 =2L +5W

if the pens are "separate" in the sense of none sharing any sides

then the max area is when you have 4 square pens

that's a total of 4x4 =16 sides

200/16 = 50/4 =25/2 = 12 1/2 by 12 1/2 feet each

each pen then has an area = 12.5x12.5 =156

4 times 156 = 624 ft^2

or, as the pens don't have to be the same size

200/4 = 50 feet for one large pen

50x50 = 2,500 ft^2 and 3 other pens of infinitesmal size nearly 0 by 0 ft

or as close to 2500 and 0 and you want

but when you say "parallel" pens that seems to mean they can share sides, so 1,000 ft^2 is the max Area

Answer 2