Expert: Socrates - 12/11/2006

Question
Courtney is hiring people to build her a pig pen. She has 150 feet of fencing to use. What dimensions should be used so that the enclosed area will be a maximum?

Answer
I assume the pen is to be a rectangle. Let L be the length and let W be the width. Then the perimeter of the fence is 2L + 2W , and since 150 feet of fencing is to be used ,
2L + 2W = 150 . So , L + W = 75 and W = 75 - L . The area , A , is A = LW . Since W = 75 - L , thus, we have
A = 75L - L^2 . To find the maximum area, set the derivative equal to 0 , A'(L) = 75 - 2L = 0 . Solving for L, 75 - 2L = 0 , so 2L = 75 and L = 37.5
Since W = 75 - L , W = 75 - 37.5 = 37.5 .

From this we see that the length and width are both 37.5 feet, so the pen is a square. The dimensions are 37.5 feet by 37.5 feet.