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Finding Critical Points of a multivariable function

  1. Jul 4, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to find (and classify, which I'm really not sure what that means) critical points of f(x,y) = x2y- x2 - 2y2 - 2x4 - 3y4

    2. Relevant equations

    3. The attempt at a solution
    I found the partial derivatives,
    fx(x,y) = 2xy - 2x - 8x3
    fy(x,y) = x2 - 4y - 12y3

    I gather that x = 0 and x = sqrt(y-1/4)
    I can't figure out how to solve for y, if I'm even supposed to.
    I was under the impression to set both equations to 0 and solve, but I don't know if I am correct.
    I guess my [STRIKE]main[/STRIKE] only question is if I am supposed to take the y solution and use it in the x? I want to solve it 100% on my own, so I only want broad answers, if that makes any sense?

    Thanks to anyone who can help me! :)
    Last edited: Jul 4, 2011
  2. jcsd
  3. Jul 4, 2011 #2
    Hi arisachu! :smile:

    You'll need to know when fx and fy simultaniously become 0. So you'll need to solve the system

    [tex]\left\{\begin{array}{c} f_x(x,y)=0\\ f_y(x,y)=0\\ \end{array}\right.[/tex]


    [tex]\left\{\begin{array}{c} 2xy - 2x - 8x^3=0\\ x^2 - 4y - 12y^3=0\\ \end{array}\right.[/tex]

    From your first equation, you have correctly deduced that x=0 or


    Now, what does that give you when you plug that in the second equation?
  4. Jul 4, 2011 #3
    Thanks for you reply and help! :D

    Ohhh, so it's as simple as that? :O
    And then I solve for [STRIKE]x[/STRIKE] y from that? :)
  5. Jul 4, 2011 #4
    Yes, solve for y in the second equation, and then you can solve for x.
  6. Jul 4, 2011 #5
    Gosh, I had been thinking you solve for y in the second equation without the first equation's x, and then set those two solutions equal to each other and I couldn't for the life of me figure it out!
    You're a life (and sanity haha) saver! Thank you so much! :smile:
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