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Maximizing a quadratic

  1. Jun 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Maximizing the function f(X)=6X-1/2X^2+82. That's it, so 6X-1/2X^2+82 = 82-1/2(X^2-12X) = 82-1/2(X-6)^2+18. So f(X)=100-1/2(X-6)^2. Since (X-6)^2 is positive for any X not =6, than the max is x=6. The part I don't understand is when rearranging the function, how do you get from -1/2(X^2-12X) to -1/2(X-6)^2+18?

    2. Relevant equations

    The focus of the problem assumes my question is already common knowledge, so I don't think there are any relevant equations.

    3. The attempt at a solution

    Not sure what to do, at first I worked backwards trying to expand (x-6)^2+18, which is X^2-12X+54, but I'm at a dead end. This isn't actually for a class, but thanks for your help in advance!
  2. jcsd
  3. Jun 6, 2008 #2
    Expand [tex]-\frac{1}{2}(x-6)^{2} [/tex] and tell us how much smaller it is than [tex]-\frac{1}{2}(x^{2}-12x)[/tex]

  4. Jun 6, 2008 #3
    alright, so -1/2(X-6)^2 expanded =-1/2X^2+6X-18. 18 smaller than -1/2(X^2-12X). Very helpful, thank you. That answers my question, but this brings me to another question. If given -1/2(X^2-12X), is there a way to use calculation to find that it equals -1/2(X-6)^2+18? I guess what I am asking is how can I look at this function, -1/2(X^2-12X) and then determine that it is equal to -1/2(X-6)^2+18? Do you factor an x and expand or is there any way to do this? Thanks again!
  5. Jun 6, 2008 #4


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    Homework Helper

    This involves a method known as completing the square. Have you learnt that?
  6. Jun 6, 2008 #5
    It is known as completing the square
    If I have some quadratic of the form say...ax^2 + bx + c then first I can factor out an a to get
    [tex]a(x^{2} +\frac{b}{a}x + \frac{c}{a})[/tex]
    Then I can add and subtract [tex] (\frac{b}{2a})^{2}[/tex] to get
    [tex]a(x^{2} +\frac{b}{a}x + \frac{b^{2}}{4a^{2}} - \frac{b^{2}}{4a^{2}} + \frac{c}{a})[/tex]
    Which can then be converted to
    [tex] a((x +\frac{b}{2a})^{2}) + d)[/tex] where [tex] d = \frac{c}{a} - \frac{b^{2}}{4a^{2}}[/tex]
    You can of course pull d out of the brackets where it would then be equal to [tex]c - \frac{b^{2}}{4a}[/tex]
    Last edited: Jun 6, 2008
  7. Jun 8, 2008 #6

    Gib Z

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    Homework Helper

    Is it part of your knowledge that the axis of symetrry of a parabola of the form y= ax^2 + bx + c is the line x=-b/2a. They introduced that to us in Australia before applying Completing the square for maxima/minima. If you do, and say the parabola is concave down - where does the maximum value always occur?
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