More Optimization — Question 5

Question 5

A person is standing at point $A$ on the shore of a straight river that is 3 km wide. Point $B$ lies on the opposite shore, 5 km downstream from the point directly opposite $A$. The person can run at 10 km/h along the shore and swim at 5 km/h directly across the river or at an angle.

The person first swims from $A$ to $P$, then runs along the opposite shore to $B$. At what point $P$ between the point directly opposite $A$ and $B$ should the person land to minimize total time to reach $B$?

Question 5 - Solution

Let the swimmer land at point $P$, which is $x$ km from the point directly across from $A$.

Step 1: Write distances:

Swim distance: $\sqrt{x^2 + 3^2} = \sqrt{x^2 + 9}$ , Run distance: $5 - x$

Step 2: Write time expression:

Swimming speed: 5 km/h , Running speed: 10 km/h

$$T(x) = \frac{\sqrt{x^2 + 9}}{5} + \frac{5 - x}{10}$$

Step 3: Minimize time — take derivative:

$$T'(x) = \frac{x}{5\sqrt{x^2 + 9}} - \frac{1}{10}$$

Set $T'(x) = 0$:

$$\frac{x}{5\sqrt{x^2 + 9}} = \frac{1}{10} \Rightarrow \frac{2x}{\sqrt{x^2 + 9}} = 1 \Rightarrow 4x^2 = x^2 + 9 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}$$

Step 4: Answer

$$\boxed{ \begin{aligned} x &= \sqrt{3} \approx 1.732\ \text{km} \\ T_{\min} &= \text{when the person lands } \sqrt{3}\ \text{km from the point directly across from } A. \end{aligned} }$$