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Question
An island is $1 \mathrm{mi}$ due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is $15 \mathrm{mi}$ west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of $6 \mathrm{mph}$ and swims at a rate of $2.5 \mathrm{mph}$. How far should the visitor run before swimming to minimize the time it takes to reach the island? Round your answer to three decimal places and omit units from the answer box.

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Step 1 ：Let's denote the point on the shore closest to the island as point A, the cabin as point B, and the point where the visitor starts swimming as point C. We know that AB = 15 miles and AC is the distance the visitor runs, so BC = 15 - AC. We also know that CA is perpendicular to AB and AC is the hypotenuse of the right triangle ABC, so by the Pythagorean theorem, AC = \(\sqrt{(BC)^2 + 1}\).

Step 2 ：The time it takes to run is AC/6 and the time it takes to swim is \(\sqrt{(BC)^2 + 1}\)/2.5. The total time is T = AC/6 + \(\sqrt{(BC)^2 + 1}\)/2.5. We want to minimize this time, so we take the derivative of T with respect to AC and set it equal to zero: \(\frac{dT}{dAC} = \frac{1}{6} - \frac{BC}{2.5\sqrt{(BC)^2 + 1}} = 0\).

Step 3 ：Solving for BC, we get BC = \(\sqrt{(\frac{2.5}{6})^2 - 1} = \sqrt{(\frac{25}{36}) - 1} = \sqrt{-\frac{11}{36}} = \frac{\sqrt{11}}{6}\) miles.

Step 4 ：However, this is not a valid solution because the square root of a negative number is not a real number. Therefore, the minimum time is achieved when BC = 0, i.e., when the visitor runs the entire 15 miles to point A and then swims 1 mile to the island.

Step 5 ：So, the visitor should run 15 miles before swimming. \(\boxed{15}\)