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Math Help - Finding two unknown numbers

  1. #1
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    Finding two unknown numbers

    Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.


    There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?
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  2. #2
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    Quote Originally Posted by brownjovi View Post
    Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.


    There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?
    x - y = -4

    x = y - 4

    P = xy = (y-4)y

    what value of y makes P a minimum?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    x - y = -4

    x = y - 4

    P = xy = (y-4)y

    what value of y makes P a minimum?
    Is it two?
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  4. #4
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    I am a newbie myself, but I think that one way to solve this is to think something along these lines.

    If y is negative, the product y(y-4) will be positive.
    If y is positive, there will be a few numbers for which the product y(y-4) is negative, and some for which the product will be positive.

    In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    If we consider ax^2+bx+c completing the square gives a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a} which when a>0 we may clearly see that a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}\geqslant c-\frac{b^2}{4a} so that it attains it's minimum at x=\frac{b}{2a}. Geometrically all we did was find the vertex of the parabola.
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  6. #6
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    Quote Originally Posted by Sabo View Post
    I am a newbie myself, but I think that one way to solve this is to think something along these lines.

    If y is negative, the product y(y-4) will be positive.
    If y is positive, there will be a few numbers for which the product y(y-4) is negative, and some for which the product will be positive.
    Actually, no. There will be an infinite wet of numbers for which the product y(y- 4) is negative. You cannot assume the number will be a whole number. Completing the square, as Drexel28 suggested, will always give you the minimum or maximum point (vertex of the parabola) of a quadratic function.

    In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.
    Again, no, there are an infinite number of them.
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  7. #7
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    Right. Stuck in integer land. =/
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