A small island is 3 miles from the nearest point P on the straight shoreline of a large lake.  If a woman on the island can row a boat 3 miles per hour and can walk 4 miles per​ hour, where should the boat be landed in order to arrive at a town 15 miles down the shore from P in the least​ time?

To find the optimal landing point for the woman to minimize her travel time, we need to set up and solve the problem using calculus and optimization techniques. Here are the steps involved:

1. **Define the variables:**

   - Let \( P \) be the point on the shoreline closest to the island.
   - Let \( A \) be the point on the island where the woman starts.
   - Let \( B \) be the point on the shoreline where she lands the boat.
   - Let \( T \) be the town 15 miles down the shore from \( P \).
   - Let \( x \) be the distance from \( P \) to \( B \).

2. **Write the distance formulas:**

   - The distance from the island to the landing point \( B \) (hypotenuse of the right triangle) is \( \sqrt{x^2 + 3^2} = \sqrt{x^2 + 9} \).
   - The distance from \( B \) to \( T \) is \( 15 - x \).

3. **Write the time formulas:**

   - The rowing time to \( B \) is \( \frac{\sqrt{x^2 + 9}}{3} \) hours (since she rows at 3 miles per hour).
   - The walking time from \( B \) to \( T \) is \( \frac{15 - x}{4} \) hours (since she walks at 4 miles per hour).

4. **Total travel time \( T(x) \):**

   \[
   T(x) = \frac{\sqrt{x^2 + 9}}{3} + \frac{15 - x}{4}
   \]

5. **Find the minimum time by differentiating \( T(x) \):**

   \[
   T'(x) = \frac{d}{dx} \left( \frac{\sqrt{x^2 + 9}}{3} + \frac{15 - x}{4} \right)
   \]

   Differentiate each term separately:

   \[
   \frac{d}{dx} \left( \frac{\sqrt{x^2 + 9}}{3} \right) = \frac{1}{3} \cdot \frac{1}{2\sqrt{x^2 + 9}} \cdot 2x = \frac{x}{3\sqrt{x^2 + 9}}
   \]

   \[
   \frac{d}{dx} \left( \frac{15 - x}{4} \right) = -\frac{1}{4}
   \]

   So:

   \[
   T'(x) = \frac{x}{3\sqrt{x^2 + 9}} - \frac{1}{4}
   \]

6. **Set the derivative equal to zero to find the critical points:**

   \[
   \frac{x}{3\sqrt{x^2 + 9}} - \frac{1}{4} = 0
   \]

   Solve for \( x \):

   \[
   \frac{x}{3\sqrt{x^2 + 9}} = \frac{1}{4}
   \]

   Cross-multiply to solve for \( x \):

   \[
   4x = 3\sqrt{x^2 + 9}
   \]

   Square both sides:

   \[
   16x^2 = 9(x^2 + 9)
   \]

   \[
   16x^2 = 9x^2 + 81
   \]

   \[
   7x^2 = 81
   \]

   \[
   x^2 = \frac{81}{7}
   \]

   \[
   x = \sqrt{\frac{81}{7}} = \frac{9}{\sqrt{7}} = \frac{9\sqrt{7}}{7}
   \]

   This is approximately:

   \[
   x \approx 3.40 \text{ miles}
   \]

7. **Verify that this value of \( x \) gives a minimum time by checking the second derivative or analyzing the behavior of the function around the critical point.**

Therefore, the woman should land the boat approximately 3.40 miles down the shoreline from point \( P \) to minimize her travel time to the town.

Would you like more details on any of these steps or have any other questions? Here are some related questions you might be interested in:

1. How do you find the critical points of a function?
2. What are the steps to differentiate a composite function?
3. How do you verify if a critical point is a minimum or maximum?
4. How does the distance formula work in coordinate geometry?
5. What are real-world applications of optimization problems?

**Tip:** When dealing with optimization problems, always double-check your derivatives and critical points to ensure accuracy.

Explore how to minimize travel time for a woman rowing and walking to a town 15 miles down the shore. Learn about optimization techniques in calculus and distance-time formulas. Suitable for advanced high school students.

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