cost=84(2W + L)
but LW=40000 to L=40000/W
cost=84(2W+40000/W)
minimum cost for a particular W will be when the derivative of cost with resjpect to W is zero...or
0=84(2 -40000/W^2) or
solve for W. Then solve for L. Then solve for cost at that L,W
A rectangular recreational field needs to be built outside of a gymnasium. Three walls of fencing are needed and the fourth wall is to be a wall of the gymnasium itself. The ideal area for such a field is exactly 40000
40000
ft2
2
. In order to minimize costs, it is necessary to construct the fencing using the least amount of material possible. Assuming that the material used in the fencing costs $84/ft, what is the least amount of money needed to build this fence of ideal area? Round your answer to the nearest two decimal places.
2 answers
You will note that the fence needs to be divided equally between lengths and widths. So, the minimum fence length is for a field that is 200√2 x 100√2
2 answers