A pig rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. He has 610 feet of fencing available to complete the job. What is the largest possible total area of the four pens?

Answer 1

two lengths and five widths ... 2 L + 5 W = 610 ... L = 305 - 2.5 W

A = L * W = 305 W - 2.5 W^2

the maximum is on the axis of symmetry of the parabola

Wmax = -b / 2a = -305 / -5 = 61

substitute back to find L ... then calculate the area

Answer 2

Well, I guess those 4 pens are to be the same size.
He could do it with 4 fences parallel with the wall
or
he could do it with 5 fences perpendicular to the wall.
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If he does it with 4 plus two ends:
A = 4 x y with y parallel to wall and length 4 x perpendicular to the wall
then
610 = 4 y + 8 x
4 y = 610 - 8 x
y = (610/4) - 2 x
then
A = 4 x [ (610/4) - 2 x]
A = 610 x - 8 x^2
that is min when dA/dx = 0 (or find vertex of that parabola)
dA/dx = 0 = 610 - 16 x
x = 610 /16 = 305/8 or about 38.125
then y = (610/4) - 2 x = (610/4) - 610/8 = 610/8
then Area = 4 x y = 4 (305/8)(610/8)= 305*610/16 = 11,628.125
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BUT the other way, 5 x lengths and one y parallel to wall
610 = 5x + y
A = x y
then A = x (610-5x) = 610 x - 5 x^2
dA/dx =0 = 610 - 10 x
x = 61
then y = 610 - 5(61) = 305
and
x y = Area = 61*305 = 18,605 THE WINNER

Answer 3

Scott - the wall so only one length

Answer 4

Damon ... don't see "wall" mentioned in the problem ...

Answer 5

There will be 2 ends and 5 cross fences. Maximum area is achieved when the fencing is divided equally among lengths and widths: 305 feet each.
So, the dimensions will be 152.5 by 61

Answer 6

You are right Scott. I misread it.