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Quick question about use of the Hessain for critical points

  1. Jun 27, 2012 #1
    So I am solving for the critical points of x3-3x-y2+9y+z2



    I've found the critical points and I'm just evaluating them in the Hessain. Now I computed the eigenvalues for the hessian, but one of my 2nd order derivatives was a constant. So my hessian looked like this (it was diagonal, so i'm only writing the diagonal terms)

    (6x - λ), (-6y-λ), (2-λ)

    Now all my C.P were of the form (±1,±√3,0)

    So whatever eigenvalues I solve for I am always going to get λ=2, but in the solutions they said that a maximum existed at (-1,√3,0) but if I solve for the eigenvalues I will have λ = -6, λ = -6√3, λ = 2. So how is this a local max if they are not all negative. (long winded for short explanation I know)
     
  2. jcsd
  3. Jun 27, 2012 #2
    Is there a typo? I get y=9/2.
     
  4. Jun 27, 2012 #3
    y2 is supposed to be y3....my bad
     
  5. Jun 27, 2012 #4
    I think you're right, they may have just over looked the z contribution, could be a typo. Notice contribution from z has no local max, so I don't think there will be a local max anywhere in ℝ3.

    Also, notice hessian is already diagonal, (6x,-6y,2), so it's not necessary to use eigenvalues to diagonalize it or find principal accelerations. In a sense, you're principal directions or eigenvectors are (dx,0,0), (0,dy,0), (0,0,dz). (Where we consider xTHx.) So x>0 and y<0 if and only if we have a local minimum (at a critical point).
     
    Last edited: Jun 27, 2012
  6. Jun 27, 2012 #5
    Ok. So that z co-ordinate should've had some sort of variable in it in order for the problem to "work out" the way it should, but for the most part I was on the right track. Thanks
     
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