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Math Help - Local extremum (2 variables)

  1. #1
    MHF Contributor arbolis's Avatar
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    Local extremum (2 variables)

    I have to find the critical points of the function f and to tell whether they are local minimum, local maximum or saddle point.
    The function is f(x,y)=x^2y^2.
    By intuition the point (0,0) is a local minimum.
    Formally, I found that f_x(x,y)=0 \Leftrightarrow 2xy^2=0, so x=0 and y= any value. (I'm not convinced since y is treated as a constant and it might be 0, while in this case x could be any other number different from 0. Hence my "so" is not an implication in fact! I don't understand what I'm doing wrong)

    f_y(x,y)=0 \Rightarrow y=0 and x= any value. (Once again I don't trust my implication)
    Hence the critical point is (0,0).

    From my class notes, if f_{xx}(0,0)f_{yy}(0,0)-f_{xxy}(0,0)=0, then I cannot conclude about the nature of the critical point. Unfortunately it happens here, and there's no theorem that can help me.
    If it was the case of a transcendantal function, I would have approximate it by a Taylor's polynomial and checking if the polynomial reaches a minimum or a maximum or a saddle point, but the problem is that the function is a polynomial. (by the way I'm not even sure that approximating a function via a Taylor's polynomial assegurate that the approximated function reaches the same critical points. Maybe there's a theorem about it but I never read it yet).
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  2. #2
    Senior Member Spec's Avatar
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    All the critical points are positive-semidefinite since Q(h,k)=2y^2h^2,\ y\in R or Q(h,k)=2x^2k^2,\ x\in R (depending on the critical point)

    Because of that, you can't use the quadratic form to see if the point is an extremum or not.

    x^2\geq 0,\ y^2\geq 0 \implies x^2y^2\geq 0

    f(x,y)=0 for all points (x,y)=(0,C),\ C\in R or (x,y)=(C,0),\ C\in R, so these points are local minima ( f'_x=0,\ f'_y=0 here as well).
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post


    From my class notes, if ?!""f_{xx}(0,0)f_{yy}(0,0)-f_{xxy}(0,0)""!?=0, then I cannot conclude about the nature of the critical point. Unfortunately it happens here, and there's no theorem that can help me.
    .
    The formula not like that but like this

    D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f^2_{xy}(x_0,y_0)
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Amer View Post
    The formula not like that but like this

    D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f^2_{xy}(x_0,y_0)
    Thanks a lot! My professor wrote the ^2 but seemed right after the _x. She didn't prove anything so I took the formula I saw, without understand why it is like it is. I miss a good book on the subject. (I can't borrow Stewart's one on vector calculus since too many people borrow it)
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