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Finding partial derivatives

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Find all solutions (x,y) for which fx(x,y) = 0 = fy(x,y) if f(x,y) = 12xy - x^2 y - 2xy^2

    2. Relevant equations



    3. The attempt at a solution

    f(x,y)=12xy-x^2y-2xy^2

    fx(x,y)=12y-2xy-2y^2
    fy(x,y)=12x-x^2-4xy

    0=12y-2xy-2y^2
    0=12x-x^2-4xy

    EQ 1: 2xy=12y-2y^2
    2x=12-2y
    x=6-y

    EQ 2: 0=12(6-y)-(6-y)^2-4(6-y)y
    0=72-12y-(y^2-12y+36)-24y+4y^2
    0=3y^2-24y+36
    0=3(y^2-8y+12)
    0=3(y-6)(y-2)

    y=6 y=2
    x=0 x=4 so (0,6) , (4,2)

    I found those 2 solutions so far, is there any more that I might have missed?
     
  2. jcsd
  3. Mar 19, 2009 #2

    lanedance

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    Homework Helper

    hi camino
    i haven't been through your working, but how about (0,0)?
     
  4. Mar 19, 2009 #3
    I thought about that, but i'm not sure how to prove that as a solution. Any idea?
     
  5. Mar 19, 2009 #4
    2xy=12y-2y^2
    2x=12-2y

    You divided by y too early instead of factoring it out. So you lost y=0 as a possible solution. Plug y=0 into both EQ1/2 and see for what x they are both equal to 0.

    I think what you did already covers the other solutions that don't have y=0.
     
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