Did you make a diagram?
Mine has I as the island, O as the point on shore opposite the island, and T as the transformer station.
So IO = 6 km and OT = 12 km
Let P be any point between O and T,
let OP = x, then PT = 12-x
Cost = 6000(IP) + 4000(12-x)
IP^2 = x^2 + 6^2
IP = (x^2 + 36)^(1/2)
Cost = 6000(x^2 + 36)^(1/2) + 48000 - 4000x
d(Cost)/dx = 3000(x^2+36)^(-1/2)(2x) - 4000
= 0 for a max/min of Cost
2x(3000)/√(x^2 + 36) = 4000
3x/√(x^2 + 36) = 2
square and cross multiply
9x^2 = 4x^ + 144
5x^2 = 144
x= √144/5 = 5.3667
Form your word conclusion
An electric utility is required to run a cable from a transformer station on the shore of a lake to an island. The island is 6 km from the shore and the station is 12 km down the shoreline from a point opposite the island. It costs $4000/km to run the cable on land and $6000/km underwater. Find the path the cable should take for a minimum cost of installation.
2 answers
say it goes 12-x km down the shoreline, then heads offshore straight for the island. The resulting path length is 12-x + the hypotenuse sqrt(36 + x^2)
cost = (12 -x)(4000) + [(36+x^2)^.5]6000
0 =dc/dx = -4000 + .5 [ (36+x^2)^-.5 ] (2x)(6000)
which is
4000 = 6000x/sqrt(36+x^2)
sqrt(36+x^2) = (3/2)x
36+x^2 = (9/4)x^2
36 = (5/4)x^2
x^2 = 28.8
x = 5.37 km
12-x = 6.63 miles on shore then head under water
cost = (12 -x)(4000) + [(36+x^2)^.5]6000
0 =dc/dx = -4000 + .5 [ (36+x^2)^-.5 ] (2x)(6000)
which is
4000 = 6000x/sqrt(36+x^2)
sqrt(36+x^2) = (3/2)x
36+x^2 = (9/4)x^2
36 = (5/4)x^2
x^2 = 28.8
x = 5.37 km
12-x = 6.63 miles on shore then head under water
8 answers