Brandon is on one side of a river that is 50 m wide and wants to reach a point 3
ID: 3192852 • Letter: B
Question
Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. Find the minimum amount of time if Brandon can swim at 2 m/s and run at 4 m/s. The minimum amount of time is ?Explanation / Answer
T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set dT/dx = 0, which will be the case when (x/2) / sqrt(2500 + x^2) = 1/6 ----> 3x = sqrt(2500 + x^2) ----> 9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, which is about 17.7 m downstream from Q. Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the minimum travel time just plug this value of x into to equation for T: T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s. =(1 minute & 13.57 seconds)