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Homework Help: Critical points.

  1. Mar 17, 2008 #1
    I have to find the critical points for the partials:

    f_x = y/3(24 - 12x - 4y) = 0
    f_y = x/3(24 - 6x - 8y) = 0

    I get y = 0 and x = (6-y)/3 for x in the first partial. How am I supposed to proceed? If I plug these into the secon, it gets nasty. Can anyone demonstrate how this is done?

    Thank you.
     
  2. jcsd
  3. Mar 17, 2008 #2

    Dick

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    First, you mean, e.g. f_x=(y/3)(24-12x-4y). Put enough parentheses in to make it unambiguous. Second, you don't get y=0 AND x=(6-y)/3, you get y=0 OR x=(6-y)/3. Plug each possibility into the second equation. Third, it doesn't get 'nasty'. You get quadratic equations for x or y. That's not considered 'nasty'.
     
  4. Mar 17, 2008 #3
    That is some gruesome math man.
     
  5. Mar 17, 2008 #4

    Dick

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    Quadratics?? Are you serious?
     
  6. Mar 17, 2008 #5
    LOL, you just made my night :)
     
  7. Mar 17, 2008 #6

    Dick

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    Well, thanks. I am trying for the comedian of the year award. Aren't you the artist formerly known as rocophysics? Why the name change?
     
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