1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook