Mixing
Extremes of the function in a certain set
$begingroup$
I need to find the extreme values of a function $$f(x,y,z)=x^2+y^2+z^2$$ on the set $S={(x,y,z) in mathbb R^3: z=xy+2}$.
Set $S$ is not compact, so we cannot be sure the local extrema exist. Guess I'd need to use differentiation, but not sure how.
Any help would be appreciated.
multivariable-calculus derivatives optimization a.m.-g.m.-inequality
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
$endgroup$
add a comment |
$begingroup$
I need to find the extreme values of a function $$f(x,y,z)=x^2+y^2+z^2$$ on the set $S={(x,y,z) in mathbb R^3: z=xy+2}$.
Set $S$ is not compact, so we cannot be sure the local extrema exist. Guess I'd need to use differentiation, but not sure how.
Any help would be appreciated.
multivariable-calculus derivatives optimization a.m.-g.m.-inequality
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
$endgroup$
$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36
add a comment |
$begingroup$
I need to find the extreme values of a function $$f(x,y,z)=x^2+y^2+z^2$$ on the set $S={(x,y,z) in mathbb R^3: z=xy+2}$.
Set $S$ is not compact, so we cannot be sure the local extrema exist. Guess I'd need to use differentiation, but not sure how.
Any help would be appreciated.
multivariable-calculus derivatives optimization a.m.-g.m.-inequality
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
$endgroup$
I need to find the extreme values of a function $$f(x,y,z)=x^2+y^2+z^2$$ on the set $S={(x,y,z) in mathbb R^3: z=xy+2}$.
Set $S$ is not compact, so we cannot be sure the local extrema exist. Guess I'd need to use differentiation, but not sure how.
Any help would be appreciated.
multivariable-calculus derivatives optimization a.m.-g.m.-inequality
multivariable-calculus derivatives optimization a.m.-g.m.-inequality
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
edited Jan 26 at 14:07
Michael Rozenberg
108k1895200
108k1895200
asked Jan 26 at 11:31
windircursewindircurse
1,154820
asked Jan 26 at 11:31
windircursewindircurse
1,154820
asked Jan 26 at 11:31
windircursewindircurse
1,154820
1,154820
$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36
add a comment |
$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36
$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36
add a comment |
1 Answer
1
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oldest
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$begingroup$
By AM-GM $$x^2+y^2+z^2=x^2+y^2+(2+xy)^2geq2|xy|+x^2y^2+4xy+4geq$$
$$geq2|xy|+x^2y^2-4|xy|+4=left(|xy|-1right)^2+3geq3.$$
The equality occurs for $|xy|=1$ and $|xy|=-xy$, id est, for $(x,y)=(1,-1)$ for example,
which says that $3$ is a minimal value of $f$.
The maximum does not exist, of course.
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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$begingroup$
By AM-GM $$x^2+y^2+z^2=x^2+y^2+(2+xy)^2geq2|xy|+x^2y^2+4xy+4geq$$
$$geq2|xy|+x^2y^2-4|xy|+4=left(|xy|-1right)^2+3geq3.$$
The equality occurs for $|xy|=1$ and $|xy|=-xy$, id est, for $(x,y)=(1,-1)$ for example,
which says that $3$ is a minimal value of $f$.
The maximum does not exist, of course.
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By AM-GM $$x^2+y^2+z^2=x^2+y^2+(2+xy)^2geq2|xy|+x^2y^2+4xy+4geq$$
$$geq2|xy|+x^2y^2-4|xy|+4=left(|xy|-1right)^2+3geq3.$$
The equality occurs for $|xy|=1$ and $|xy|=-xy$, id est, for $(x,y)=(1,-1)$ for example,
which says that $3$ is a minimal value of $f$.
The maximum does not exist, of course.
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By AM-GM $$x^2+y^2+z^2=x^2+y^2+(2+xy)^2geq2|xy|+x^2y^2+4xy+4geq$$
$$geq2|xy|+x^2y^2-4|xy|+4=left(|xy|-1right)^2+3geq3.$$
The equality occurs for $|xy|=1$ and $|xy|=-xy$, id est, for $(x,y)=(1,-1)$ for example,
which says that $3$ is a minimal value of $f$.
The maximum does not exist, of course.
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
By AM-GM $$x^2+y^2+z^2=x^2+y^2+(2+xy)^2geq2|xy|+x^2y^2+4xy+4geq$$
$$geq2|xy|+x^2y^2-4|xy|+4=left(|xy|-1right)^2+3geq3.$$
The equality occurs for $|xy|=1$ and $|xy|=-xy$, id est, for $(x,y)=(1,-1)$ for example,
which says that $3$ is a minimal value of $f$.
The maximum does not exist, of course.
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 12:53
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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$begingroup$
This is a typical usage of Lagrange's multipliers. You should look for some reference on it but to make it short, every extremum $X = (x,y,z)$ on $S$ will satisfy $Df_X = lambda Dg_X$ and $g(X) = 0$ where $g$ is the function defining your set $g(x) = xy + 2 - z$ (it is a necessary condition, not a sufficient one). So solve this equation and check each solution you find one by one. PS : D is the differential, you can replace it by the gradiant and this will work the same
$endgroup$
– Junkyards
Jan 26 at 12:10
$begingroup$
I do not think Lagrange multiplier is at all needed , just substitute z in the expression to get an explicit formula for f in terms of x,y and then make the gradient zero (one thing to check is whether the function is differentiable but in this case this is not an issue as it can be seen easily the function is differentiable)
$endgroup$
– Bijayan Ray
Jan 27 at 6:36