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Math Help - Application of Maxima and Minima (Unresolved Problem)

  1. #1
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    Application of Maxima and Minima (Unresolved Problem)

    Im having difficulties solving for x in this problem:


    a lighthouse is at point A, 4km offshore from the nearest point O of a straight beach; a store is at point B, 4km down the beach from O. if the lighthouse keeper can row 4km/hr & walk 5km/hr, how should he proceed in order to get from the lighthouse to the store in the least possible time?


    Given: Distance from A to O : 4 km
    Distance from O to B : 4 km
    Constant Speed (to row) : 4 kmph
    Constant Speed (to walk) : 5 kmph
    Let C - a point between O and B
    x - distance from O&C
    (4-x) - distance from C&B
    sqrt(16+x^2) - distance from A to C

    Speed = Distance / Time hence T = Distance / Speed

    Time = Time (row) + Time (walk)
    = ( distance from A to C / Constant Speed to row ) + ( distance from C to B / Constant Speed to walk )
    = ( sqrt(16+x^2) / 4 ) + ( (4-x)/5)

    T' = ( x / 4sqrt(16+x^2) ) - ( 1/5 )

    Equating T' = 0 results to x = 16 /3

    since OB > OC where OB = 4 km and OC < 4

    then x cannot be 16/3 because 16/3 > 4 !

    HELP!!! im so confused!!!!

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  2. #2
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    Re: Application of Maxima and Minima (Unresolved Problem)

    General theorem- a continuous function always has a max and min on a closed, bounded interval.
    The max and min must occur at one of three kinds of points- where the derivative is 0, where the derivative is not defined, or at the endpoints.

    What you have done is show that that there is NO point where the derivative is 0 inside the interval , nor is there any point where the derivative is not defined, so the max and min must be at the endpoints.
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  3. #3
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    Re: Application of Maxima and Minima (Unresolved Problem)

    Quote Originally Posted by HallsofIvy View Post
    General theorem- a continuous function always has a max and min on a closed, bounded interval.
    The max and min must occur at one of three kinds of points- where the derivative is 0, where the derivative is not defined, or at the endpoints.

    What you have done is show that that there is NO point where the derivative is 0 inside the interval , nor is there any point where the derivative is not defined, so the max and min must be at the endpoints.
    =======

    So if the minimum occurs at the endpoints, does it mean the lighthouse keeper has to row a distance of 4 sqrt(2) km heading the store (directly) which gives him a minimum time = sqrt(2) h ~ 1.414 h ???
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  4. #4
    Grand Panjandrum
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    Re: Application of Maxima and Minima (Unresolved Problem)

    Quote Originally Posted by teachmemath247 View Post
    =======

    So if the minimum occurs at the endpoints, does it mean the lighthouse keeper has to row a distance of 4 sqrt(2) km heading the store (directly) which gives him a minimum time = sqrt(2) h ~ 1.414 h ???
    Yes.

    CB
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