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Finding Critical Points and Local Extrema of a Multivariable Function

  1. Feb 20, 2006 #1
    For f(x,y) = x^2 + y^2 + 3xy I need to find the critical points and prove whether or not they are local minima, maxima or saddle points. I thought the only critical point was (0,0) since Df = (2x + 3y, 2y + 3x) = 0. Doesn't this make (0,0) a local min? The reason I doubt this now is because upon constructing the Hessian matrix and finding the eigenvalues, I got lambda = 5 and -1 (which would show that the point is a saddle point). Can someone please help me with this?
     
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  3. Feb 20, 2006 #2

    0rthodontist

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    Why are you finding the eigenvalues? What you want to do with the Hessian is find the determinant. The point is a local minimum.
     
  4. Feb 20, 2006 #3
    For the problem I am also supposed to find the eigenvalues of the Hessian. Anyway, I think the Hessian is 2 3,3 2 (first row, second row) which makes the determinant 4-9 or -5. This means the det(Hf(x,y)) < 0, which makes the critical point a saddle (second derivative test for local extrema). Is it really a local min?
     
  5. Feb 20, 2006 #4

    0rthodontist

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    No, I am wrong, you are right. It's a saddle point.
     
  6. Feb 20, 2006 #5

    benorin

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    Notice that [tex]f(x,y)=x^2 + y^2 + 3xy =(x+y)^2+xy[/tex]

    so apply the change of variables:

    [tex] u=x+y, v=x-y [/tex]

    which has as its inverse

    [tex]x=\frac{1}{2}(u+v),y=\frac{1}{2}(u-v)[/tex]

    to get

    [tex]f\left( \frac{1}{2}(u+v), \frac{1}{2}(u-v)\right) =u^2+\frac{1}{4}(u+v)(u-v) = \frac{5}{4}u^2-\frac{1}{4}v^2 [/tex]

    which clearly has no extrema (neither local nor global) and the only ciritical point is the saddle point u=v=0, this corresponds to x=y=0 which is likewise a saddle point for the Jacobian of the transformation is

    [tex]J = \left| \frac{\partial (u,v)}{\partial (x,y)}\right| = \left| \begin{array}{cc} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{array}\right| = \left| \begin{array}{cc} 1&1\\ 1&-1\end{array}\right| = 1\cdot 1 - 1\cdot (-1) = 2[/tex]

    and the transformation is then just a rotation and a contraction of the old variables.
     
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