If the width is w, the length is 250000/w, so the perimeter is
p = 2w + 250000/w
dp/dw = 2 - 250000/w^2
dp/dw=0 when w = 500/√2 = 353.6 ft
So, the least fencing needed is 1414.2 ft
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 250000ft2. 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 $53/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
along building = w
out from building = L
total length = w + 2 L = p
area = A = w L =250,000
so L = 250,000/w
then the fencing perimeter
p = w + 2 L = w + 500,000/w
dp/dw = 0 for minimum
dp/dw = 1 - 500,000/w^2 = 0
w = sqrt (500,000) = 707.1
L = 353.6
p = w + 2 L = 1414.2
cost = 1414.2*53 = 74,952.96
out from building = L
total length = w + 2 L = p
area = A = w L =250,000
so L = 250,000/w
then the fencing perimeter
p = w + 2 L = w + 500,000/w
dp/dw = 0 for minimum
dp/dw = 1 - 500,000/w^2 = 0
w = sqrt (500,000) = 707.1
L = 353.6
p = w + 2 L = 1414.2
cost = 1414.2*53 = 74,952.96
2 answers