The problem is to minimize the total time taken for the man to reach point $B$, which involves a combination of rowing and running. Here's how we can approach solving it:

Variables

1. The river is 1 km wide (perpendicular distance from $A$ to $C$).
2. The destination $B$ is 1 km downstream from $C$.
3. Rowing speed: $6 \, \text{km/h}$.
4. Running speed: $8 \, \text{km/h}$.
5. We need to find the optimal landing point $D$, which is a distance $x$ downstream from $C$, so that total time is minimized.

Total Time Function

The total time $T$ is the sum of:

• Time to row from $A$ to $D$: $T_{\text{row}} = \frac{\sqrt{1^2 + x^2}}{6}$
• Time to run from $D$ to $B$: $T_{\text{run}} = \frac{1 - x}{8}$ Thus: $T(x) = \frac{\sqrt{1 + x^2}}{6} + \frac{1 - x}{8}$

Optimization

To minimize $T(x)$, we differentiate it with respect to $x$ and find where $\frac{dT}{dx} = 0$.

1. Differentiate $T(x)$: $\frac{dT}{dx} = \frac{x}{6\sqrt{1 + x^2}} - \frac{1}{8}$
2. Set $\frac{dT}{dx} = 0$ and solve for $x$: $\frac{x}{6\sqrt{1 + x^2}} = \frac{1}{8}$ Multiply through by $6\sqrt{1 + x^2}$: $8x = 6\sqrt{1 + x^2}$
3. Square both sides to eliminate the square root: $64x^2 = 36(1 + x^2)$ Simplify: $64x^2 = 36 + 36x^2$ $28x^2 = 36$ $x^2 = \frac{36}{28} = \frac{9}{7}$ $x = \sqrt{\frac{9}{7}} = \frac{3}{\sqrt{7}} \approx 1.13 \, \text{km}$

Interpretation

The man should land approximately 1.13 km downstream from point $C$ to minimize the time to reach $B$.

Would you like a detailed explanation of any step or numerical confirmation of the result? Here are some related questions you might find useful:

1. How do we calculate the rowing time in similar situations?
2. How does running speed influence the result if it changes?
3. What would happen if the river's width changes?
4. How do we account for a current in the river's speed?
5. Can we solve this using numerical methods for greater precision?

Tip: For real-world optimization problems, always check