A pig farmer wants to enclose a rectangular area and then divide it into three pens with fencing parallel to one side of the rectangle (see the figure below). There are 940 feet of fencing available to complete the job. What is the largest possible total area of the three pens?

Answer 1

let the length of the large rectangle be y
let each of the parallel divider sides be x
so we have 4x + 2y = 940
2x + y =470
y = 470 - 2x

Area of whole field
= xy
= x(470-2x)
= 470x - 2x^2
d(Area)/dx = 470 - 4x
= 0 for a max of Area
4x = 470
x = 117.5
then y = 235
and the largest area = xy = 117.5(235) = 27612.5 ft^2

If you don't know Calculus, then find the vertex of the downwards opening parabola
Area = -2x^2 + 470x

the x of the vertex is -b/2a = -470/-4 = 117.5
then proceed as above