Expert: Scotto - 3/26/2011

Question
A farmer has a rectangular pen, which he divides into 5 identical pig pens. Each of the five pig pens has an area of 2400 m squared. The dimensions must be atleast 10m. What is the lesat amount of fencing he can use to create the pig pens?

Answer
The best set up would be to put the 5 pens in a row.
For the sides together, which I will refer to as x, they will be 6x of fencing used,
for there will be one fence between each pair, for 4x, plus 1 on either end, making 6x.

For the sides that are on either end of each paster, which I will refer to as y,
the total will be 10y, for there are 5(2y) since there are two for each pastor.

There area is A(x,y) = 5xy.

It is also known that the area of each is 2400, and that is xy.
That means y = 2400/x.  Putting this in the perimeter equation of 2x + 2y, we get the perimeter to be p(x) = 2x + 4800/x.

Now that we have a formula for t he perimeter in terms of x, take the derivative and get
p'(x) = 2 - 4800/x^2.  Setting this to 0 gives 0 = 2 - 4800/x^2, and that can be changed
to 2 = 4800/x^2, so x^2 = 2400.  This makes x aprroximately 49, for 49^2 = 2,401.

Since y = 2400/x and x is almost 50, y is roughly 50.
In actuallity, x = y = 48.9899, but that is so close to 50 I'll just call it 50.

In actuality, call it 48.98979486 on each side and that maximizes the area in my spreadsheet calculations.


The total area of each pen would then be (50 m) x (120 m) = 6000 m^2.

The total length of fence needed is 10x + 6y.