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