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Math Help - optmization - crossing river with moving water

  1. #1
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    optmization - crossing river with moving water

    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:
    total time = time to cross river+time to cross land
    time on river = \frac{1}{3}\cos{\theta} = \frac{1}{3}\sec{\theta}
    OP=(12+3\sin{\theta}))\frac{1}{3}\sec{\theta} = \frac{\sin{\theta}+4}{\cos{\theta}}
    PB=10-OP=\frac{5}{3}-\frac{\frac{\sin{\theta}}{6}+\frac{2}{3}}{\cos{\th  eta}}
    total time = \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 -\frac{\frac{\sin{\theta}}{3}+\frac{1}{6}}{\cos{\th  eta}^2} and set this equal to 0. Then solving for θ I get \frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6}

    problem: none of these angles make sense becaue they need to be between \frac{\pi}{2} and 0.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The pitcure is a little hard to interpretate, but the 'spontaneous' quetion is: why the solution \theta= -\frac{\pi}{6} is not acceptable? ...

    Kind regards

    \chi \sigma
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  3. #3
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    So both \frac{-5\pi}{6} and \frac{-\pi}{6} work
    Last edited by superdude; October 29th 2009 at 10:48 AM. Reason: completely changed
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