Find the extremas of $f(x,y,z)=xyz-x^2-y^2-z^2$
$begingroup$
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.
calculus multivariable-calculus hessian-matrix
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
$endgroup$
add a comment |
$begingroup$
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.
calculus multivariable-calculus hessian-matrix
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
$endgroup$
add a comment |
$begingroup$
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.
calculus multivariable-calculus hessian-matrix
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
$endgroup$
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
calculus multivariable-calculus hessian-matrix
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
asked Dec 21 '18 at 12:25
J. DoeJ. Doe
14213
14213
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
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.
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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votes
$begingroup$
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.
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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.
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 21 '18 at 12:29
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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