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Math Help - optimization

  1. #1
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    optimization

    find the point on the curve y=x^1/2 nearest to the point (c,0) if c(> and =)1/2
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joyce View Post
    find the point on the curve y=x^1/2 nearest to the point (c,0) if c(> and =)1/2
    We need to minimize the distance between the points (x, x^{1/2}) and (c, 0).

    The distance between two the two points is:
    d = \sqrt{(x - c)^2 + (x^{1/2} - 0)^2}

    Note that if we square the distance:
    d^2 = (x - c)^2 + (x^{1/2} - 0)^2
    we still find the minimum x in the same place with a much easier computation.

    So now take your x derivative with respect to the d^2 equation, set it equal to 0, etc.

    Not sure why you've got the c \geq 1/2 part, though.

    Edit: Ahhhh, c \geq 1/2 because the domain of the function is only positive x.

    -Dan
    Last edited by topsquark; October 4th 2007 at 07:53 AM. Reason: I'll get this right yet!
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