A small island is 4 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 13 miles down the shore from $P$ in the least time?
The boat should be landed Fniles down the shore from $P$. (Type an exact answer.)
Answer
Therefore, the boat should be landed $\boxed{\frac{12\sqrt{7}}{7}}$ miles down the shore from $P$ to minimize the total time.
Step 1 ：Given a small island is 4 miles from the nearest point $P$ on the straight shoreline of a large lake. A woman on the island can row a boat 3 miles per hour and can walk 4 miles per hour. We need to find where should the boat be landed in order to arrive at a town 13 miles down the shore from $P$ in the least time.
Step 2 ：Let's denote the distance from the point $P$ to the landing point by $x$. Then the distance from the landing point to the town is $13 - x$.
Step 3 ：The time taken to row the boat is the distance divided by the speed, which is $\frac{\sqrt{x^2 + 4^2}}{3}$ (using the Pythagorean theorem to find the distance from the island to the landing point).
Step 4 ：The time taken to walk to the town is $\frac{13 - x}{4}$.
Step 5 ：So the total time taken is $T(x) = \frac{\sqrt{x^2 + 4^2}}{3} + \frac{13 - x}{4}$.
Step 6 ：To find the minimum of this function, we can take the derivative and set it equal to zero. The derivative of $T(x)$ is $\frac{x}{3\sqrt{x^2 + 16}} - \frac{1}{4}$.
Step 7 ：Solving for $x$ gives the critical point $x = \frac{12\sqrt{7}}{7}$. This is the only critical point in the domain of the problem ($0 \leq x \leq 13$), so it must be the point that minimizes the total time.
Step 8 ：To confirm this, we can check the second derivative of the function at this point. If the second derivative is positive, then the function has a local minimum at the critical point.
Step 9 ：The second derivative of $T(x)$ is $-\frac{x^2}{3(x^2 + 16)^{3/2}} + \frac{1}{3\sqrt{x^2 + 16}}$. The second derivative at the critical point is $\frac{7\sqrt{7}}{768}$, which is positive.
Step 10 ：Since the second derivative at the critical point is positive, the function has a local minimum at this point. Since this is the only critical point in the domain of the problem, it must be the global minimum.
Step 11 ：Therefore, the boat should be landed $\boxed{\frac{12\sqrt{7}}{7}}$ miles down the shore from $P$ to minimize the total time.
