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 9 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 from the refinery should p be located?

  7.85 km

Step-by-step explanation:

Let x represent the distance from the refinery to point P. Then the distance under the river from point P to the storage tanks is ...

  √(2² +(9 -x)²) = √(x² -18x +85)

In units of $200,000, the cost of the pipeline will be ...

  c = x + 2√(x² -18x +85)

The derivative with respect to x is ...

  dc/dx = 1 +(2x -18)/√(x² -18x +85)

When we set this to zero, we can get the equation ...

  √(x² -18x +85) +(2x -18) = 0

And this can be rewritten as ...

  3x^2 -54x +239 = 0

which has solution

  x = 9 -(2/3)√3 ≈ 7.8453

Point P should be located about 7.85 km from the refinery.

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It is interesting to note that the angle the pipe makes with the riverbank is given by arccos(1/2), where the 1/2 is the ratio of overland to under-river costs. That angle is 60°, so the distance along the riverbank from the storage facility to point P is 2cot(60°) = (2/3)√3 ≈ 1.1547 km. You will recognize this as the value subtracted from 9 km in the solution above.

This "angle solution" is the generic solution to this sort of problem where the route costs are different and part of the route is along the edge of the higher-cost path.

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