Question

A farmer has 160 feet of fencing to enclose 2 adjacent rectangular pig pens. What dimensions should be used so that the enclosed area will be a maximum?

Answer 1

I'm assuming that the pig pens have identical dimensions.

Explanation:

Let's assume that the pig pens need to be fenced in the way shown in the diagram above.

Then, the perimeter is given by `4x + 3y = 160`.

`4x = 160 - 3y`

`x = 40 - 3/4y`

The area of a rectangle is given by `A = L xx W`, however here we have two rectangles put together, so the total area will be given by `A = 2 xx L xx W`.

`A = 2(40 - 3/4y)y`

`A = 80y - 3/2y^2`

Now, let's differentiate this function, with respect to y, to find any critical points on the graph.

`A'(y) = 80 - 3y`

Setting to 0:

`0 = 80 - 3y`

`-80 = -3y`

`80/3 = y`

`x = 40 - 3/4 xx 80/3`

`x = 40 - 20`

`x = 20`

Hence, the dimensions that will give the maximum area are `20` by `26 2/3` feet.

A graphical check of the initial function shows that the vertex is at `(26 2/3, 1066 2/3)`, which represents one of the dimensions that will give the maximum area and the maximum area, respectively.

Hopefully this helps!