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Minimum Point of a Function

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Decide whether F=x2y2-2x-2y has a minimum at the point x=y=1 (after showing that the first derivatives are zero at that point).

    2. Relevant equations

    FxxFyy-Fxy2

    3. The attempt at a solution

    So I found that:
    Fx=2xy2-2, which at point (1,1) = 0 OK
    Fy=2x2y-2, which at point (1,1) = 0 OK
    Fxx=2y2
    Fyy=2x2
    Fxy=4xy

    So at point (1,1):
    FxxFyy-Fxy2=4-16=-12, which is less than 0.

    Does this mean that because it is less than zero, rather than greater, it DOES NOT have a minimum but rather a saddle point? Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 30, 2011 #2

    I like Serena

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    Homework Helper

    Hi tatianaiistb! :smile:

    Yes, the second derivative test says that a negative determinant means it's a saddle point.

    Btw, you can easily verify by checking a few points in the neighbourhood, like (1,0) and (-1.1).
     
  4. Nov 30, 2011 #3

    Ray Vickson

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    Homework Helper

    You get a saddle point. To understand WHY (rather than just citing prescriptions), look at what you get when you use the Hessian matrix to construct a quadratic approximation near (1,1): f ~ 2(x-1)^2 + 2(y-1)^2 -8(x-1)(y-1) = 2[(x-1)-(y-1)]^2-4(x-1)(y-1) = 2(x-y)^2 -4(x-1)(y-1) = 2(u-v)^2-4uv, where u=x-1 and v=y-1. When v=0 we have f ~ 2u^2 > 0 for nonzero u; when u=0 we have f~2v^2 > 0 for nonzero v. But when u=v we have f ~ -4u^2 < 0 for nonzero u (and v), so (u,v) = (0,0) is a min along some direction lines and a max along other direction lines. That is what a saddle point means.

    RGV
     
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