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Find stationary points of a two variable function involving

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all stationary points of the function

    G(x, y) = (x^3)*e^(−x^2−y^2)

    2. Relevant equations
    fx=0 and fy=0

    3. The attempt at a solution

    Gx = 3x^2*e^(-x^2-y^2) +x^3(-2x)e^(-x^2-y^2) = e^(-x^2-y^2)(3x^2-2x^4)

    Gx = 0 implies 3x^2-2x^4=0

    x^2(3-2x^2)=0

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

    Gy = (-2y)(x^3)(e^(-x^2-y^2))

    Gy=0

    implies (-2y)(x^3)(e^(-x^2-y^2))=0

    y=0 x = 0

    (+or- (3/2)^(1/2), 0) are the two stationary points according to wolfram alpha but I don't understand why (o,a), where a is a real number, isn't a solution since Gx=Gy=0 when x = 0.

    Any help would be appreciated.
     
    Last edited by a moderator: Nov 24, 2015
  2. jcsd
  3. Nov 24, 2015 #2

    Ray Vickson

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    The points (0,a) are solutions, but (according to you---I have not checked) Wolfram Alpha seems to have missed them. Maple did not miss them.
     
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