To solve this problem, we need to determine where the man should land in order to reach point B as quickly as possible.

1. Rowing to point D:

   Let's assume the point D is $x$ km downstream from C. This means:

   • The distance he rows (AD) can be calculated using the Pythagorean theorem: $\text{AD} = \sqrt{(4\, \text{km})^2 + x^2}$

   • The time taken to row to D: $t_{\text{row}} = \frac{\sqrt{16 + x^2}}{6}$

   • The distance he runs (DB) is $1 - x$ km.

   • The time taken to run to B: $t_{\text{run}} = \frac{1 - x}{8}$

   • The total time $t$ is: $t = \frac{\sqrt{16 + x^2}}{6} + \frac{1 - x}{8}$

2. Minimizing the Total Time:

   We want to find the value of $x$ that minimizes $t$.

   To do this, take the derivative of $t$ with respect to $x$ and set it to zero: $\frac{d}{dx} \left( \frac{\sqrt{16 + x^2}}{6} + \frac{1 - x}{8} \right) = 0$

   Differentiating, we get: $\frac{x}{6\sqrt{16 + x^2}} - \frac{1}{8} = 0$

   Solving for $x$, you get: $\frac{x}{6\sqrt{16 + x^2}} = \frac{1}{8}$

   Cross-multiply and simplify: $8x = 6\sqrt{16 + x^2}$

   $64x^2 = 36(16 + x^2)$

   $64x^2 = 576 + 36x^2$

   $28x^2 = 576$

   $x^2 = \frac{576}{28}$

   $x^2 = 20.57$

   $x \approx \sqrt{20.57} \approx 4.53$

3. Conclusion: According to the calculations, point D is approximately 0.53 km downstream from C. Therefore, to minimize the time to reach B, he should land about 0.53 km downstream from C.