Let's solve this step by step.

Problem Breakdown

1. The river is 70 meters wide, and the target point is 250 meters downstream.
2. Brandon swims diagonally across the river at 2 m/s and runs on land at 4 m/s.
3. We need to find the minimum time required to reach the target.

Key Variables and Concepts

• Let Brandon swim a diagonal path to some point along the opposite riverbank.
• The diagonal swimming path creates a right triangle with:
  • The width of the river (70 m) as one leg.
  • The horizontal distance covered while swimming as part of the hypotenuse.

Let the point where Brandon exits the water on the opposite side be $x$ meters downstream from his starting position. Then:

• The remaining distance he runs on land is $(250 - x)$.

Formulating the Total Time

The total time $T$ is the sum of:

1. Time spent swimming: $T_{\text{swim}} = \frac{\sqrt{x^2 + 70^2}}{2}$ (Hypotenuse distance divided by swimming speed).

2. Time spent running: $T_{\text{run}} = \frac{250 - x}{4}$

So, the total time $T$ becomes: $T(x) = \frac{\sqrt{x^2 + 70^2}}{2} + \frac{250 - x}{4}$

Minimize $T(x)$

To find the minimum time, we differentiate $T(x)$ with respect to $x$ and solve for when the derivative equals zero.

1. Differentiate $T(x)$: $\frac{dT}{dx} = \frac{x}{2\sqrt{x^2 + 70^2}} - \frac{1}{4}$

2. Set $\frac{dT}{dx} = 0$: $\frac{x}{2\sqrt{x^2 + 70^2}} = \frac{1}{4}$

3. Solve for $x$: Multiply through by $4\sqrt{x^2 + 70^2}$: $2x = \sqrt{x^2 + 70^2}$

   Square both sides: $4x^2 = x^2 + 4900$

   Simplify: $3x^2 = 4900 \quad \Rightarrow \quad x^2 = \frac{4900}{3} \quad \Rightarrow \quad x = \sqrt{\frac{4900}{3}} \approx 40.41 \, \text{m}.$

Calculate the Total Time

Using $x = 40.41$:

1. Hypotenuse (swimming distance): $\sqrt{x^2 + 70^2} = \sqrt{(40.41)^2 + 70^2} \approx \sqrt{4900} = 70 \, \text{m}.$

2. Time swimming: $T_{\text{swim}} = \frac{70}{2} = 35 \, \text{s}.$

3. Time running: $T_{\text{run}} = \frac{250 - 40.41}{4} = \frac{209.59}{4} \approx 52.4 \, \text{s}.$

4. Total time: $T = 35 + 52.4 = 87.4 \, \text{s}.$

Final Answer

The minimum time is 87.40 seconds.

Let me know if you want more clarification or steps explained further!

Questions for Expansion:

1. How is the hypotenuse of the swimming distance calculated geometrically?
2. What if Brandon’s swimming speed increases—how does it affect the time?
3. How do we derive the critical point using calculus principles?
4. Why is $T(x)$ minimized using the derivative?
5. Could alternative paths result in a similar or shorter time?

Tip:

Always double-check units (meters, seconds, etc.) in optimization problems to ensure consistency!