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 5 m/s

Answer 1

Let P be Brandon's starting point and Q be the point directly across the river from P.
Now let R be the point where Brandon swims to on the opposite shore, and let
QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run
a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is

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.



mind blown