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Critical points.

  • Thread starter frasifrasi
  • Start date
  • #1
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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.
 

Answers and Replies

  • #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'.
 
  • #3
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That is some gruesome math man.
 
  • #4
Dick
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Quadratics?? Are you serious?
 
  • #5
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Quadratics?? Are you serious?
LOL, you just made my night :)
 
  • #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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