A farmer decides to make three identical pens with 88 feet of fence. The pens will be next to each other sharing a fence and will be up against a barn. The barn side needs no fence. What dimensions for the total enclosure (rectangle including all pens) will make the area as large as possible?



Answer :

Answer:

The long fence, AC = 44 feet

4 shorter segments at 11feet each = 44 feet

Area = 484 ft^2

Step-by-step explanation:

The three pens will share one fence section parallel to the barn, we'll call it AB.  Each pen will need two side fence segments, from the front fence to the barn.  We'll call those portions AC.  We need four of them:  two for the outer sides and two more to divide up the space equally among the three pens.  The total area would be AB*AC.

We know that AC + 4AB = 88 feet.

or AC = (88feet - 4AB)

Area = AB(88feet - 4AB)

Area = 88AB - 4AB^2

To maximize the area, we can take the first derivative and set it equal to zero:

Area' = 88 - 8AB

0 = 88 - 8AB

AB = 11 feet

AC = (88 - 4AB)

AC = (88 - 44)

AC = 44 feet

The fours segments of fence:  

AC + 4AB = 88 feet

(44) + 4(11 feet) = 88feet

Area = 44' x 11 ft' = 484 ft^2