A farmer has 336 feet of fencing to enclose 2 adjacent rectangular pig pens sharing a common side. What dimensions should be used for each pig pen so that the enclosed area will be a maximum? The two adjacent pens have the same dimensions.

x = 84 ft the largest side of the pens

y = 56 ft the common side is that length

Step-by-step explanation:

Let call x the total length of one side, and y the other side, and the common side such as:

p (perimeter) = 336 = 2x + 3y y = (336 - 2x) / 3

And the area of the whole area

A (t) = x * y A (x) = x * (336 - 2x) / 3 A (x) = [336x - 2x² ]/3

Taking derivatives both sides of the equation

A' (x) = [ (336 - 4x) * 3]/9 ⇒ A' (x) = 0 (336 - 4x) * 3 = 0

336 - 4x = 0

x = 336/4

x = 84 ft and y = (336 - 2x) / 3 y = (336 - 168) / 3

y = 56 ft