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Old 04-15-2010   #2
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
Originally Posted by superhume View Post
What would be the approach for this problem:
A power house, P, is on one bank of a straight river 200m wide, and a factory, F, is on the other bank 400m downstream from P. The cable has to be taken across the river under water at a cost of $12.00/m. On land the cost is $6.00/m. What path should be chosen so the cost is minimized?

Any help would be much appreciated.
W = distance under water
L = distance on land

Cost = 12*W + 6*L

If the cable is run underwater in a straight line to the opposite bank to some point between the factory and a point straight across from the power house, then the distance underwater is the hypotenuse of the right triangle with legs 200m and 400-L meters.

W^2 = 200^2 + (400-L)^2
W = sqrt(40000 + 160000 - 800*L + L^2)
W = sqrt(200000 - 800*L + L^2)

Now plug W back into the cost equation.

Cost = 12*sqrt(200000 - 800*L + L^2) + 6*L

One way to find the minimum cost is to take the derivative of cost set it equal to 0, and solve for L.

d/dL(cost) = (6*sqrt(200000 - 800*L + L^2) + 12*L - 4800)/sqrt(200000 - 800*L + L^2)

Setting the der of cost to 0 gives:
0 = 6*sqrt(200000 - 800*L + L^2) + 12*L - 4800
(4800 - 12*L)/6 = sqrt(200000 - 800*L + L^2)
Squaring both sides gives:
(144*L^2 - 115200*L + 23040000)/36 = L^2 - 800*L + 200000
36*L^2 - 28800*L + 7200000 = 144*L^2 - 115200*L + 23040000
0 = 108*L^2 - 86400*L + 15840000

Using quadratic formula gives L = 284.53 or L = 515.47

Throw away the value greater than 400 and plug the value for L back into the W question to solve for W.

W = sqrt(40000 + (400 - 284.53)^2) = 230.94

Min cost = $4478.46. The cable should be run to a point 284.53m from the factory on the opposite bank of the river.
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