Question

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $200,000/km over land to a point P on the north bank and $400,000/km under the river to the tanks. To minimize the cost of the pipeline, how far (in km) from the refinery should P be located? (Round your answer to two decimal places.)

Answer 1

As usual, draw a diagram. Let
R = refinery
T = tank farm
Q = point directly across the river from T
x = the distance RP
So QT = 2
The land distance is x
underwater distance is z^2 = (6-x)^2 + 2^2
Now the cost (in $100,000s) of the pipeline is
c(x) = 2x + 4√((6-x)^2 + 4)
now just find where dc/dx = 0
x = 6 - 2/√3

Note that as usual in these problems, the angle θ made by the pipeline with the coastline is such that sinθ is the ratio of the cost on land to that in water
In this case, tanθ = 2/4 = 1/2. You note that PQ/QT = 1//3

Answer 2

My sketch has R as the refinery on the north side, S as the location of the
storage tanks on the south side, and Q as the point directly across the river from S. Let PQ = x, then PR = 6-x, and QS = 2

Path of pipeline = SP + PR

Since we are not asked for the actual cost, only the path route, and the fact
that I noticed that the costs of laying the pipe are simply in the ratio of 2 : 1,
we can say:
Cost = 2SP + 1PR
also SP^2 = x^2 + 2^2
we have
Cost = 2(x^2 + 4)^(1/2) + 1(6-x)
dCost/dx = (x^2 + 4)^(-1/2)(2x) - 1 = 0 for a min of Cost
2x/√(x^2 + 4) = 1
2x = √(x^2 + 4)
square both sides
4x^2 = x^2 + 4
3x^2 = 4
x = 2/√3 =1.16 km
so 6-x = 4.85

They should aim for a point 4.85 km from the refinery