Question

A zoo exhibit is to be made by: enclosing a rectangular area, and then divided horizontally down the middle perpendicular to one side, then finally dividing it vertically 4 times perpendicular to the other side. They will be using 3,000 feet of fencing. (This forms ten pens.) What are the dimensions that will maximize the area enclosed? For credit, you must show how to use Calculus to find the dimensions, and justify that you have a maximum.

Answer 1

Let total length is 5a, and width is 2b, hence every pens have dimension of length = a, width = b

Total fencing length in terms of a and b,

3 horizontal length of 5a required

whereas 6 vertical length of 2b is required.

Therefore, total fencing length = 3*5a + 6*2b = 3000 feet.

=> 5a + 4b = 1000

=> b = (1000 - 5a) / 4

Area of each pens = ab

We need to maximized enclosed are of each pens.

Thus , Area in terms of a = A = a*(1000 - 5a)/4

differentiating both side, and equating it to zero will give critical point. (as it is lower bounded by zero area , thus at critical point we will get maximum area. There is no need to for 2nd derivative to check maxima or minima.)

for critical point,

Thus length of each pens is a = 100, and b = 125 (in feet)

or total length of rectangular area is 5a = 500 feet, and 2b = 250 feet ___(Answer)