- #1
Glissando
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Hey guys I know this questions has been posted before but I still can't seem to figure it out. Thanks a lot for your help (:
A power line is to be constructed from a town to a mine that is located near a straight road going to the town. The point on the road closest to the mine is 7.5km from the town and 3km from the mine. The cost of building along the road is 4500$ per km, and through the forest to the mine, 7500$ per km. Find the minimum cost of the power line.
Cost = Distance of road * 4500 + Distance through forest *7500
C = Road(4500) + Forest (7500)
C = (7.5-x)4500 + 7500sqrt(x^2+9)
dC/dx = -4500x + .5(x^2+9)^-.5(2x)(7500)
dC/dx = -4500x + 7500x(x^2+9)^-.5
dC/dx = 500x [-9 + 15(x^2+9)^-.5]
0 = 500x[-9 + 15(x^2+9)^-.5]
0 = -9 + 15(x^2+9)^-.5
9/15 = (x^2+9)^-.5
x = -0.36
Thanks in advance (:
Homework Statement
A power line is to be constructed from a town to a mine that is located near a straight road going to the town. The point on the road closest to the mine is 7.5km from the town and 3km from the mine. The cost of building along the road is 4500$ per km, and through the forest to the mine, 7500$ per km. Find the minimum cost of the power line.
Homework Equations
Cost = Distance of road * 4500 + Distance through forest *7500
The Attempt at a Solution
C = Road(4500) + Forest (7500)
C = (7.5-x)4500 + 7500sqrt(x^2+9)
dC/dx = -4500x + .5(x^2+9)^-.5(2x)(7500)
dC/dx = -4500x + 7500x(x^2+9)^-.5
dC/dx = 500x [-9 + 15(x^2+9)^-.5]
0 = 500x[-9 + 15(x^2+9)^-.5]
0 = -9 + 15(x^2+9)^-.5
9/15 = (x^2+9)^-.5
x = -0.36
Thanks in advance (: