Question

An island is 3 mi due north of its closest point along a straight shoreline. You are staying at a cabin on the shore that is 7 mi west of that point. You are planning to go from the cabin to the island. Suppose you run at a rate of 9 mph and row at a rate of 5 mph. How far should you run (in miles) before rowing to minimize the time it takes to reach the island

Answer 1

Answer:

  5.00 miles

Step-by-step explanation:

Let x represent the distance in miles you should run. Then the distance you will be rowing is ...

  √(3² +(7-x)²)

and your total travel time in hours is ...

  t = x/9 + √(3² +(7-x)²)/5

This is minimized when its derivative with respect to x is zero.

  $\dfrac{dt}{dx}=0=\dfrac{1}{9}+\dfrac{-2(7-x)}{5\cdot 2\sqrt{3^2+(7-x)^2}}\\\\0=\dfrac{5\sqrt{x^2-14x+58}+9(x-7)}{45\sqrt{x^2-14x+58}}$

This will be zero when the numerator is zero, so ...

  $0 = 5\sqrt{x^2-14x+58}+9(x-7)\\\\25(x^2-14x+58)=81(x^2-14x+49) \quad\text{subtract 9(x-7) and square}\\\\56x^2-784x+2519=0 \quad\text{write in standard form}$

Solving this quadratic by your favorite method gives ...

  x ≈ 4.996 . . . . . . there is an extraneous solution at x ≈ 9

You should run 5.00 miles before rowing in order to minimize the time to reach the island.

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Generic solution

For travel speeds a and b, where a < b and b represents the speed along the shore, the distance from the point nearest the island is given by tan(arcsin(a/b)) times the distance to the island.

Here, that is (3 mi)(tan(arcsin(5/9)) ≈ 2.004 miles. Since we're running from a point 7 miles from the point nearest the island, our running distance is 7 -2.004 = 4.996 miles.

If the starting point is less than the distance computed above, then the shortest time path is a straight line to the island.

In short, the travel angle from a line perpendicular to shore is given by arcsin(a/b).

This same solution works for problems involving laying pipeline, walking through woods, or any other scenario where there is an optimal straight-line path to a point where the cost changes.