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 can swim at the rate of 5 ft/sec and can... 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 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 Variables Let ( x ) be the distance (in feet) the lifeguard swims downstream on the opposite bank. The total distance downstream to the person is 100 ft, thus the distance he will run after swimming is ( 100 - x ).

2. Calculate Swimming Distance Using the Pythagorean theorem, the distance the lifeguard swims across the river (width) is 50 ft, and the downstream distance is ( x ). The swimming distance ( d_s ) is calculated as: $$ d_s = \sqrt{x^2 + 50^2} $$

3. Time to Swim The time taken to swim across the river ( T_s ) is given by: $$ T_s = \frac{d_s}{\text{swimming speed}} = \frac{\sqrt{x^2 + 50^2}}{5} $$

4. Time to Run The time taken to run to the person ( T_r ) is given by: $$ T_r = \frac{100 - x}{\text{running speed}} = \frac{100 - x}{15} $$

5. Total Time The total time ( T ) as a function of ( x ) is: $$ T(x) = T_s + T_r = \frac{\sqrt{x^2 + 50^2}}{5} + \frac{100 - x}{15} $$

6. Minimize Total Time To find the point that minimizes time, take the derivative of ( T(x) ) and set it to zero: $$ T'(x) = \frac{1}{5} \cdot \frac{x}{\sqrt{x^2 + 50^2}} - \frac{1}{15} $$

7. Solve for ( x ) Set the derivative ( T'(x) = 0 ) and solve for ( x ): $$ \frac{1}{5} \cdot \frac{x}{\sqrt{x^2 + 50^2}} = \frac{1}{15} $$

8. Cross Multiply and Rearrange Cross-multiply to simplify: $$ 3x = \sqrt{x^2 + 50^2} $$ Square both sides: $$ 9x^2 = x^2 + 2500 $$ Rearranging yields: $$ 8x^2 = 2500 $$

9. Find ( x ) Solve for ( x ): $$ x^2 = \frac{2500}{8} = 312.5 $$ $$ x = \sqrt{312.5} \approx 17.68 \text{ ft} $$