Pre Calculus
posted by Lauren .
A contractor is to fence off a rectangular field along a straight river, the side along the river requiring no fence.
What is the least amount of fencing needed to fence of 30,000 meters squared?

perimeter = L + 2 W = p
L W = 30,000
so
L = 30,000/W
then
p = 30,000/W + 2 W
dp/dW = 30,000/W^2 + 2
for in or max, dp/dW = 0
30,000 = 2 W^2
W^2 = 15,000
W = 122.5
then L = 30,000/122.5 = 244.9 
above gives p = 244.9+2*122.5 = 490
If you do not know derivatives then
p = 30,000/W + 2 W
2 W^2  Wp +30,000 = 0
W^2  (p/2)W = 15,000
W^2  (p/2)W + p^2/16 = 15,000 + p^2/16
(Wp/4)(Wp/4) = 1/16( p^2  240,000)
that is vertex at W = p/4 and p = sqrt(240,000) or 490 sure enough
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