Question

An oil refinery is located 1 km north of the north bank of a straight river that is 3 km wide. A pipeline is to be constructed from the refinery to storage tankslocated on the south bank of the river 9 km east of the refinery. The cost of laying pipe is $200,000 per km over land to a point P on the north bank and $400,000 perkm under the river to the tanks. To minimize the cost of the pipeline, how far down river from the refinery should the point P be located?

Answer 1

Let R = refinery, S = storage tanks, Q = point on south bank opposite P.
Let RP = x, then, QS = 3 - x, and we already know PQ = 1.

Using Pythagoras in triangle PQS, PS^2 = 1^2 + (3 - x)^2, so PS = √(x^2 - 6x + 10).

Total distance of pipeline is therefore: D = x + √(x^2 - 6x + 10).

Total cost is thus: C = 200000x + 400000√(x^2 - 6x + 10).

To minimise this, take the derivative and set it equal to zero.
dC/dx = 200000 + 400000(1/2)[1/√(x^2 - 6x + 10)](2x - 6) = 0

Simplifying gives: 3x^2 - 18x + 26 = 0
Solving gives: x = (9 ± √3)/3
Discard the positively signed solution as this gives x > 3 which is invalid.

Therefore, x = (9 - √3)/3 ≈ 2.42 km