A lifeguard wishes to get a person located 100 ft downstream on the opposite bank of a 50 ft wide river, as fast as possible. The lifeguard plans to get a person by a combination o... A lifeguard wishes to get a person located 100 ft downstream on the opposite bank of a 50 ft wide river, as fast as possible. The lifeguard plans to get a person by a combination of swimming across the river to a point x ft downriver on the opposite bank and running the rest of the way along the river. The lifeguard can swim at the rate of 5 ft/sec and can run at a rate of 15 ft/sec. To what point on the opposite bank should the lifeguard swim to get the person in the least amount of time? Read more

Steps to Solve

1. Define the Variables

Let ( x ) be the distance downstream on the opposite bank where the lifeguard swims to. The distance across the river is fixed at 50 ft.

2. Calculate the Swimming Distance

The swimming distance, ( d_s ), from the starting point to the point ( x ) on the opposite bank can be found using the Pythagorean theorem: $$ d_s = \sqrt{50^2 + x^2} $$

3. Calculate the Time Taken to Swim

The time taken to swim across, ( t_s ), is given by: $$ t_s = \frac{d_s}{\text{swimming speed}} = \frac{\sqrt{50^2 + x^2}}{5} $$

4. Calculate the Distance to Run

The distance the lifeguard runs downstream after swimming is ( (100 - x) ) feet.

5. Calculate the Time Taken to Run

The time taken to run, ( t_r ), is: $$ t_r = \frac{100 - x}{15} $$

6. Total Time Function

The total time function, ( T(x) ), to get the person is: $$ T(x) = t_s + t_r = \frac{\sqrt{50^2 + x^2}}{5} + \frac{100 - x}{15} $$

7. Minimize the Total Time Function

To find the optimal point, differentiate ( T(x) ) with respect to ( x ) and set the derivative to zero: $$ T'(x) = \frac{1}{5} \cdot \frac{x}{\sqrt{50^2 + x^2}} - \frac{1}{15} $$ Set ( T'(x) = 0 ): $$ \frac{1}{5} \cdot \frac{x}{\sqrt{50^2 + x^2}} = \frac{1}{15} $$

8. Solve for ( x )

Cross-multiply and simplify the equation: $$ 3x = \sqrt{50^2 + x^2} $$ Squaring both sides gives: $$ 9x^2 = 2500 + x^2 $$ Rearranging leads to: $$ 8x^2 = 2500 $$ Thus, $$ x^2 = \frac{2500}{8} $$ Taking the square root results in: $$ x = \frac{50}{\sqrt{2}} = 25\sqrt{2} \approx 35.36 \text{ ft} $$