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Multivariable Calculus: finding relative extrema/saddle points

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I was given an assignment to find all relative extrema and saddle points of the equation

    f(x,y) = 1/3x^4 + 1/2y^4 - 4xy^2 +2x^2 + 2y^2 + 3

    I derived the first partial with respect to x and the first partial with respect to y, but when I tried to find where they both equal 0, the problem became really complicated. I don't know what I am suppose to do. Rather stuck.

    2. Relevant equations

    partial x: 4/3x^3 - 4y^2 + 4x

    0 = 1/3x^3 - y^2 + x

    partial y: 2y^3 - 8xy + 4y

    0 = 2y(y^2 - 4x +2)

    3. The attempt at a solution

    I have tried using substition after setting the equations equal to each other, by using the partial derivatives to find

    x = (y^2 + 2)/4


    y = + or - sqrt(4x - 2)

    y = + or - sqrt(1/3x^3 + x)

    I have tried setting these two equal to each other to solve for x, but I get a cubic that I don't know what to do with. Is there something like "completing the cube" I could use? Haha.

    (1/3x^3 - 3x + 2 = 0)

    I'm so confused...please help me. :(
  2. jcsd
  3. Feb 23, 2010 #2
    For starters, (0,0) is clearly a critical point.
  4. Feb 23, 2010 #3
    I found that one, at the very least. :)

    The assignment was due today, posting this was a last hurrah, I suppose. Now, however, I'm merely curious as to how to solve it.
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