Question:

A man stands at a point A on the bank of a straight river 1km wide which is flowing at speed of 12km/h. To reach point B, 10km down the river on the other side, he first rows the boat to a point P on the opposite side and then walks the remaining distance to B. He can row at a speed of 3 km/h (in still water) and he walks 6km/h. The river flows at 12 km/h.

Determine the angle θ that would get the man to point B the fastest.Work:

Let O be the place where the boat starts. Consider point P as if it were moving towards O at 12km/h. This makes the water stationary.

equations:

$\displaystyle total time = time to cross river+time to cross land$

$\displaystyle time on river = \frac{1}{3}\cos{\theta} = \frac{1}{3}\sec{\theta}$

$\displaystyle OP=(12+3\sin{\theta}))\frac{1}{3}\sec{\theta} = \frac{\sin{\theta}+4}{\cos{\theta}}$

$\displaystyle PB=10-OP=\frac{5}{3}-\frac{\frac{\sin{\theta}}{6}+\frac{2}{3}}{\cos{\th eta}}$

total time = $\displaystyle \frac{1}{3}\sec{\theta}+\frac{5}{3}-\frac{\frac{\sin{\theta}{6}}+\frac{2}{3}}{\cos{\th eta}}$

I then take the derivative of this which is $\displaystyle -\frac{\frac{\sin{\theta}}{3}+\frac{1}{6}}{\cos{\th eta}^2}$ and set this equal to 0. Then solving for θ I get $\displaystyle \frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6}$

problem: none of these angles make sense becaue they need to be between $\displaystyle \frac{\pi}{2}$ and 0.