Question

A farmer has 20 yards of fencing to build a pen for her chickens. She decides to use a side of her barn as one side of the fenced-in area. What is the maximum area she can achieve?

Answer 1

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.

Explanation:

Given that farmer shall construct a rectangular fenced-in area and a side of the barn is one side of such area, the needed length of fencing is represent by the following perimeter equation (
`p`), measured in square yards:

`p = 2\cdot l + w`

Where:

`l` - Length of the rectangle, measured in yards.

`w` - Width of the rectangule (side of the barn), measured in yards.

In addition, the equation of the fenced-in area (
`A`) is:

`A = w\cdot l`

If
`p = 20\,yd`, equation of area is now simplified as follows:

`A = (20\,yd - 2\cdot l)\cdot l`

`A = 20\cdot l - 2\cdot l^(2)`

The value of
`l` associated with the maximum area is obtained with the help of First and Second Derivative Tests. Firstly, first and second derivatives of the area function are determined:

`A' = 20 - 4\cdot l`

`A'' = -4`

Let equalize first equation to zero, second derivative indicates that critical value follows to an absolute maximum. Hence:

`20-4\cdot l = 0`

`l = 5\,yd`

The width of the rectangle is: (
`p = 20\,yd` and
`l = 5\,yd`)

`w = p - 2\cdot l`

`w = 20\,yd - 2\cdot (5\,yd)`

`w = 10\,yd`

And finally, the maximum area she can achieve is:

`A = (5\,yd)\cdot (10\,yd)`

`A = 50\,yd^(2)`

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.