Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$












0












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Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$.





After some calculation, $Df(x,y,z)=0$ for and only for $$
\p_1=(0,0,0), p_2=(2,2,2),p_3=(-2,-2,2),p_4=(-2,2,-2),p_5=(2,-2,-2)
\ H(f)=begin{vmatrix}
-2 &z &y \
z & -2 &x \
y & x &-2
end{vmatrix}
$$

So this test helps only for the maxima $p_1$. About the other critical points, I've tried to calculate $f(x,y,z)-f(x+varepsilon ,y,z)$ but both for positive and negative $varepsilon$ the sign remains the same.










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    0












    $begingroup$


    Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$.





    After some calculation, $Df(x,y,z)=0$ for and only for $$
    \p_1=(0,0,0), p_2=(2,2,2),p_3=(-2,-2,2),p_4=(-2,2,-2),p_5=(2,-2,-2)
    \ H(f)=begin{vmatrix}
    -2 &z &y \
    z & -2 &x \
    y & x &-2
    end{vmatrix}
    $$

    So this test helps only for the maxima $p_1$. About the other critical points, I've tried to calculate $f(x,y,z)-f(x+varepsilon ,y,z)$ but both for positive and negative $varepsilon$ the sign remains the same.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$.





      After some calculation, $Df(x,y,z)=0$ for and only for $$
      \p_1=(0,0,0), p_2=(2,2,2),p_3=(-2,-2,2),p_4=(-2,2,-2),p_5=(2,-2,-2)
      \ H(f)=begin{vmatrix}
      -2 &z &y \
      z & -2 &x \
      y & x &-2
      end{vmatrix}
      $$

      So this test helps only for the maxima $p_1$. About the other critical points, I've tried to calculate $f(x,y,z)-f(x+varepsilon ,y,z)$ but both for positive and negative $varepsilon$ the sign remains the same.










      share|cite|improve this question









      $endgroup$




      Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$.





      After some calculation, $Df(x,y,z)=0$ for and only for $$
      \p_1=(0,0,0), p_2=(2,2,2),p_3=(-2,-2,2),p_4=(-2,2,-2),p_5=(2,-2,-2)
      \ H(f)=begin{vmatrix}
      -2 &z &y \
      z & -2 &x \
      y & x &-2
      end{vmatrix}
      $$

      So this test helps only for the maxima $p_1$. About the other critical points, I've tried to calculate $f(x,y,z)-f(x+varepsilon ,y,z)$ but both for positive and negative $varepsilon$ the sign remains the same.







      calculus multivariable-calculus hessian-matrix






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      asked Dec 21 '18 at 12:25









      J. DoeJ. Doe

      14713




      14713






















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          Take the point $p_2$. Then your matrix is$$begin{bmatrix}-2&2&2\2&-2&2\2&2&-2end{bmatrix},$$whose eigenvalues are $-4$ (twice) and $2$. So, $p_2$ is a saddle point of $f$. Apply the same approach to the other critical points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
            $endgroup$
            – J. Doe
            Dec 21 '18 at 12:32








          • 1




            $begingroup$
            No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
            $endgroup$
            – José Carlos Santos
            Dec 21 '18 at 12:33












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          $begingroup$

          Take the point $p_2$. Then your matrix is$$begin{bmatrix}-2&2&2\2&-2&2\2&2&-2end{bmatrix},$$whose eigenvalues are $-4$ (twice) and $2$. So, $p_2$ is a saddle point of $f$. Apply the same approach to the other critical points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
            $endgroup$
            – J. Doe
            Dec 21 '18 at 12:32








          • 1




            $begingroup$
            No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
            $endgroup$
            – José Carlos Santos
            Dec 21 '18 at 12:33
















          1












          $begingroup$

          Take the point $p_2$. Then your matrix is$$begin{bmatrix}-2&2&2\2&-2&2\2&2&-2end{bmatrix},$$whose eigenvalues are $-4$ (twice) and $2$. So, $p_2$ is a saddle point of $f$. Apply the same approach to the other critical points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
            $endgroup$
            – J. Doe
            Dec 21 '18 at 12:32








          • 1




            $begingroup$
            No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
            $endgroup$
            – José Carlos Santos
            Dec 21 '18 at 12:33














          1












          1








          1





          $begingroup$

          Take the point $p_2$. Then your matrix is$$begin{bmatrix}-2&2&2\2&-2&2\2&2&-2end{bmatrix},$$whose eigenvalues are $-4$ (twice) and $2$. So, $p_2$ is a saddle point of $f$. Apply the same approach to the other critical points.






          share|cite|improve this answer









          $endgroup$



          Take the point $p_2$. Then your matrix is$$begin{bmatrix}-2&2&2\2&-2&2\2&2&-2end{bmatrix},$$whose eigenvalues are $-4$ (twice) and $2$. So, $p_2$ is a saddle point of $f$. Apply the same approach to the other critical points.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 12:29









          José Carlos SantosJosé Carlos Santos

          174k23133243




          174k23133243












          • $begingroup$
            If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
            $endgroup$
            – J. Doe
            Dec 21 '18 at 12:32








          • 1




            $begingroup$
            No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
            $endgroup$
            – José Carlos Santos
            Dec 21 '18 at 12:33


















          • $begingroup$
            If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
            $endgroup$
            – J. Doe
            Dec 21 '18 at 12:32








          • 1




            $begingroup$
            No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
            $endgroup$
            – José Carlos Santos
            Dec 21 '18 at 12:33
















          $begingroup$
          If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
          $endgroup$
          – J. Doe
          Dec 21 '18 at 12:32






          $begingroup$
          If $H(f)$ at a point $p$ has 2 same eigenvalues then $p$ is a saddle point? @jos
          $endgroup$
          – J. Doe
          Dec 21 '18 at 12:32






          1




          1




          $begingroup$
          No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
          $endgroup$
          – José Carlos Santos
          Dec 21 '18 at 12:33




          $begingroup$
          No. It's bacause at least one eigenvalue is positice, at least one eigenvalue is negative and $0$ is not an eigenvalue.
          $endgroup$
          – José Carlos Santos
          Dec 21 '18 at 12:33


















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