A farmer has 144 feet of fencing to use to create a rectangular enclosure for his pigs. He decides to use the wall of his barn as one side of the enclosure, as shown. Note: picture not drawn to scale The farmer uses function A to model the area of the enclosure, in square feet, when it is x feet wide. What width should the farmer use to make the enclosure in order to have the maximum area for his pigs? The best form of the equation for finding the required information is (standard, vertex or factored) . The farmer should make the enclosure (48, 36, 72 or 144) feet wide to have the maximum area for his pigs.

The question tests the applications of differential calculus. We are informed that the farmer will use the wall of his barn as one side of the enclosure implying that the 144 feet of fencing will only be used to form the three remaining sides of the rectangular enclosure.

If we let x denote the width of this rectangular enclosure to be formed, then its length will be 144-2x since we shall be having two sides each measuring x and the total length of the fencing is 144. 2x will be used on the width leaving 144-2x for the length since one side of the length is formed by the wall.

If we let A denote the area of the rectangular enclosure, then;

A = x(144 - 2x)

since the area of a rectangle is simply the product of the length and the width. We open the brackets;

A = 144x - 2x^2

Clearly, the Area is a function of the width that is A is the dependent variable while x is the independent one.

At the maximum area, the slope of the tangent to the curve 144x - 2x^2 is 0. Therefore, we differentiate the given function of x with respect to x then set the resulting expression to 0 to determine x that will maximize A;

dA/dx = 144 - 4x

We set this expression to 0 and solve for x;

144 - 4x = 0

144 = 4x

x = 36

Therefore, the farmer should make the enclosure 36 feet wide to have the maximum area for his pigs.