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Math Help - the nearest point sort of problems

  1. #1
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    the nearest point sort of problems

    It is a minimization problem and i don't really understand them. Please give me some reasonable solution on the following one:

    Find the point (
    x, y) belong toR2 on the graph of y = x^(1\2) nearest to the point (4, 0).

    Thanks X 100000
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  2. #2
    Super Member

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    Hello, saskadimova!

    Find the point (x,y) on the graph of y\,=\,x^{\frac{1}{2}} nearest to the point (4, 0).
    Perhaps a sketch will help . . .
    Code:
            |
            |             *
            | (x,y) *
            |   o
            | *  \
            |*    \d
            |      \
        - - * - - - o- - - - -
            |     (4,0)

    The point has coordinates (x,y) \,=\,\left(x,\sqrt{x}\right)

    We want its distance from (4,0) to be a minimum.

    The distance is: . d \;=\;\sqrt{(x-4)^2 + \left(\sqrt{x} - 0\right)^2}

    . . So we have: . d \;=\;\left(x^2-7x+16\right)^{\frac{1}{2}}

    Then: . d\,' \;=\;\tfrac{1}{2}\left(x^2-7x+16\right)^{-\frac{1}{2}}(2x-7) \quad\Rightarrow\quad \frac{2x-7}{2\sqrt{x^2-7x+16}} \;=\;0

    Hence: . 2x-7\:=\:0 \quad\Rightarrow\quad x \:=\:\frac{7}{2}

    And: . y \:=\:\sqrt{\frac{7}{2}} \:=\:\frac{\sqrt{14}}{2}


    Therefore, the nearest point is: . \left(\frac{7}{2},\:\frac{\sqrt{14}}{2}\right)

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