Question

brandon is on one side of a river that is 50 m wide and wants to reach a point 200 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. brandon can swim at 1.5 m/s and run at 4 m/s.
(a) let ( t(x) ) be the total time brandon takes to swim and then run along the river. determine ( t(x) ).
3
(b) determine the distance ( x ) so that the total time is optimized.

Explanation:

Part (a)

Step1: Define Variables and Swim Distance

Let \( x \) be the distance Brandon runs downstream. The river is 50 m wide, so the swimming distance (by Pythagoras) is \( \sqrt{x^2 + 50^2}=\sqrt{x^2 + 2500} \) meters. Time = Distance / Speed.

Step2: Time for Swimming

Swim speed is 1.5 m/s, so swim time \( t_{swim}=\frac{\sqrt{x^2 + 2500}}{1.5} \).

Step3: Time for Running

Run distance is \( 200 - x \) (since total downstream is 200 m), run speed is 4 m/s, so run time \( t_{run}=\frac{200 - x}{4} \).

Step4: Total Time Function

Total time \( T(x)=t_{swim}+t_{run}=\frac{\sqrt{x^2 + 2500}}{1.5}+\frac{200 - x}{4} \). Simplify \( \frac{1}{1.5}=\frac{2}{3} \), so \( T(x)=\frac{2\sqrt{x^2 + 2500}}{3}+\frac{200 - x}{4} \).

Step1: Differentiate \( T(x) \)

First, find \( T'(x) \). The derivative of \( \frac{2\sqrt{x^2 + 2500}}{3} \) is \( \frac{2}{3}\cdot\frac{2x}{2\sqrt{x^2 + 2500}}=\frac{2x}{3\sqrt{x^2 + 2500}} \). The derivative of \( \frac{200 - x}{4} \) is \( -\frac{1}{4} \). So \( T'(x)=\frac{2x}{3\sqrt{x^2 + 2500}}-\frac{1}{4} \).

Step2: Set Derivative to Zero

To optimize, set \( T'(x) = 0 \):
\( \frac{2x}{3\sqrt{x^2 + 2500}}=\frac{1}{4} \)
Cross - multiply: \( 8x = 3\sqrt{x^2 + 2500} \)
Square both sides: \( 64x^2 = 9(x^2 + 2500) \)
Expand: \( 64x^2=9x^2 + 22500 \)
Subtract \( 9x^2 \): \( 55x^2 = 22500 \)
\( x^2=\frac{22500}{55}=\frac{4500}{11}\approx409.09 \)
\( x=\sqrt{\frac{4500}{11}}\approx\frac{30\sqrt{5}}{\sqrt{11}}\approx\frac{30\times2.236}{3.317}\approx20.2 \) (we take positive root as distance can't be negative).

Step3: Verify it's a Minimum

We can check the second derivative or test intervals. The second derivative test (or knowing it's a distance - time optimization with a single critical point) confirms this is a minimum.

Answer:

\( T(x)=\frac{2\sqrt{x^2 + 2500}}{3}+\frac{200 - x}{4} \)

Part (b)