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Function in 3 variables, determinant of the Hessian=0

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    find the minima and maxima of the following function:
    ##f:\mathbb{R}^3 \to \mathbb{R} : f(x,y,z)=x(z^2+y^2)-yx##

    3. The attempt at a solution

    after computing the partials, i see ∇f=0 for every point in the x-axis: (a, 0, 0)
    The Hessian is:
    ( 0 0 0 )
    ( 0 2a -1 )
    ( 0 -1 2a )
    for every value of a, the determinant is 0.
    the eigenvalues are: ##\lambda_1=0##
    ##\lambda_{2,3}= 2a±\sqrt{1}##=##2a\pm1 ##
    ##\lambda_{2,3}## are both positive if a>1/2 and both negative if a<-1/2.
    Thus, for -1/2<a<1/2 i can say the points are inflection points, because the eigenvalues have oppisite sign.
    But how about ##a \geq 1/2## and ## a \leq -1/2##? i get two positive/negative eigenvalues, but the first one is always zero, so i can't really say that they are maxima/minima, right? what's the method to determine if they are max/min? thank you very much
     
    Last edited: Nov 16, 2013
  2. jcsd
  3. Nov 16, 2013 #2
    Is it correct to say that, being semidefinite positive for ##x \geq1/2## the function is convex and has therefore infinite minima and being semidefinite negative for ##x \leq -1/2## it is concave and so all the points to the left of (-1/2, 0, 0) are maxima?
     
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