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Thread: Hard Q, Help Now

  1. #1

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    Hard Q, Help Now

    I need help please. Thankyou.

    A bushwalker can walk at 5km/h through clear land and 3km/h through bushland. If she has to het from point a to point b following a route indicated in the image, find the value of x so that the route is covered in a minimum time. (Note: time = distance/speed)

    Image: http://img294.imageshack.us/my.php?i...ntitledtp0.jpg

    Please help, i need helkp getting to the answer. Please post all working. thnks

  2. #2
    vbuggy krtxmrtz's Avatar
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    Re: Hard Q, Help Now

    Welcome to the forums.

    I'll give you the general solution and then will substitute the actual values. Let A and B the horizontal and vertical distances respectively from a to b. Call v1 and v2 the walking velocities in the bush and clear respectively and t1 and t2 the times taken to cross each region.

    The total time to be minimized is
    T = t1 + t2 = x1 / v1 + x1 / v2

    Now, x1 = Sqr(a2 + x2) and x2 = b - x. Then,

    T = Sqr(a2 + x2) / v1 + (b - x) / v2

    For the x value that minimizes T, the derivative of T with respect to x must be 0:

    0 = dT/dx = -1/v2 + x/v1Sqr(a2 + x2)

    From here and with some algebra,

    x = a/Sqr([v2/v1]2 - 1)

    Finally,

    v1 = 3 km/h
    v2 = 5 km/h
    a = 2 km
    b = 3 km

    This yields: x = 1.5 km
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