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Math Help - Maximum value

  1. #1
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    Maximum value

    Why does: n/(n+k) have a maximum value of 1 for k = 0, and for no other values of k?
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  2. #2
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    Quote Originally Posted by Aquafina View Post
    Why does: n/(n+k) have a maximum value of 1 for k = 0, and for no other values of k?
    Hi

    n being fixed, as k increases, (n+k) increases, 1/(n+k) decreases, n/(n+k) decreases
    The maximum value is therefore obtained for k=0
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Now if n were fixed, like my friend runnig gag says, with k being a variable. this function has no limit as to how high it will go.

    When the denominator (n+k) aproaches zero, The function grows without bound and therefore has no maximum height.

    Is n fixed or not? Because this would change things a bit.
    Last edited by VonNemo19; May 1st 2009 at 06:02 PM. Reason: 1
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  4. #4
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    no max.

    considering it be a function of k,

    its a rectangular hyperbola,

    we have an asymptote at k=-n,

    so there is basically no max. (it goes up to infinity)
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  5. #5
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    thanks..

    in reply to the first post, how is k=0 the maximum value the expression is decreasing for 1/n+k and n/n+k like you said?

    and with the rectangular hyperbole, so that would imply no maximum value? i.e. it is minus infinity? but we assuming k is = to or > 0 so we take 0?
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  6. #6
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    k=0 is not the max value.

    and the rectangular hyperbola(y=1/x) looks like so:



    your would just shift left/right as per the transformation. so the max/min values remain the same.

    it takes all values between -infinity to +infinity
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Hey, that's what i said!
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    Hey, that's what i said!
    ??
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  9. #9
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    hi, thank, no n is not fixed, it is basically a probability:

    n = number of balls
    k = how many balls are withdrawn from a bag

    and in a particular situation the probability comes out to: n/n+k
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