Asked 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?
What is the least amount of fencing needed to fence of 30,000 meters squared?
Answers
Answered by
Damon
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
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
Answered by
Damon
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
(W-p/4)(W-p/4) = 1/16( p^2 - 240,000)
that is vertex at W = p/4 and p = sqrt(240,000) or 490 sure enough
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
(W-p/4)(W-p/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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