Optimization Problem

Question

An island is 10 mi from the nearest point on a straight shoreline; that point is 8 mi from a power station. A utility company plans to lay electrical cable underwater from the island to the shore and then underground along the shore to the power station. Assume that is costs $3000 per mile to lay cable underwater and $2000 per mile to lay underground cable. At what point should the underwater cable meet the shore to minimize the cost of the project.

Answer 1

Choose the point, O, on the shore directly opposite the island as zero. Suppose the cable meets the short at point A, which is x miles from the zero and then travels along the X-axis to the power station.

Let D(x) = distance from the island to A.

Then, by Pythagoras: D(x) = √(10² + x²) = √(x² + 100)

Cost to lay this cable = C(x) = 3 D(x) + 2 (8 - x) . . . . . . . . . . working in $,000

= 3 √(x² + 100) + 2 (8 - x) . . . . . . . . . . . . . . (1)

C'(x) = 3(1/2) (2x) (x² + 100)^-1/2 - 2

We have min or max cost when C'(x) = 0

=> 3 (1/2) 2x (x² + 100)^-1/2 - 2 = 0

=> 3x / (x² + 100)^1/2 = 2

3x = 2 (x² + 100)^1/2 . . . . . . . multiplying by (x² + 100)^1/2

9x² = 4 (x² + 100) . . . . . . . . . squaring both sides

5x² = 400

x² = 80

x = √80 = 8.9 miles

The low cost answer is for the cable to come ashore 0.9 mile beyond the power station! Most unexpected.

The least cost is found by plugging x = 8.9 into (1)