A farmer has 100 yards of fencing to form two identical rectangular pens

Question

and a third pen that is twice as long at the other two pens, as shown in the
diagram at right. All three pens have the same width, x. What value of y produces a maximum total fenced area?

Answer 1

You don't say how the pens are configured. I will assume that the long pen is adjacent to the other two, and they share a long side. So, if the pens have dimensions x*y, x*y and x*2y then the fencing needed is
5x+4y = 100
the area is
A = 2xy + x*2y = 4xy = 4x(100-5x)/4 = x(100-5x)
dA/dx = 100-10x
max area when x=10, so y = 50/4

As with all these problems, maximum area is achieved when the fencing is divided equally among lengths and widths.

Answer 2

Oobleck's solution is wrong. The amount of fencing needed would actually be 5x + 6y = 100. His method to solve is still correct, though.

5x + 6y = 100
x = -(6/5)y+20

A = 2xy
A = 2y[-(6/5)y+20] = -12y^2/5 + 40y

A' = -24y/5 +40
24y = 200
y = 25/3