Finding extrema of a tri-variable function under a constraint
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We need to find extremum of $$f(x,y,z) = yz$$ under the constraint $$g(x,y,z) = 2x^2 + 3y^2 + z^2 - 12xy + 14xz - 35$$
Using the technique of Lagrange Multipliers, leads to four simultaneous equations which are quite tedious to solve esp. in an exam-setup under limited duration of time.
Entirely alternate solutions and/or techniques that efficiently solves the above system of equations are welcome.
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
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show 2 more comments
$begingroup$
We need to find extremum of $$f(x,y,z) = yz$$ under the constraint $$g(x,y,z) = 2x^2 + 3y^2 + z^2 - 12xy + 14xz - 35$$
Using the technique of Lagrange Multipliers, leads to four simultaneous equations which are quite tedious to solve esp. in an exam-setup under limited duration of time.
Entirely alternate solutions and/or techniques that efficiently solves the above system of equations are welcome.
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
$endgroup$
1
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So it is $f(x, y, color{red}{z})$? or what is $z$?
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– caverac
Jan 13 at 18:47
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My fault; apologies:( Editing in a moment.
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– Winged Blades of Godric
Jan 13 at 18:48
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Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
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Is the constraint $g(x,y,z)=0$, then?
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– amd
Jan 13 at 20:44
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55
|
show 2 more comments
$begingroup$
We need to find extremum of $$f(x,y,z) = yz$$ under the constraint $$g(x,y,z) = 2x^2 + 3y^2 + z^2 - 12xy + 14xz - 35$$
Using the technique of Lagrange Multipliers, leads to four simultaneous equations which are quite tedious to solve esp. in an exam-setup under limited duration of time.
Entirely alternate solutions and/or techniques that efficiently solves the above system of equations are welcome.
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
$endgroup$
We need to find extremum of $$f(x,y,z) = yz$$ under the constraint $$g(x,y,z) = 2x^2 + 3y^2 + z^2 - 12xy + 14xz - 35$$
Using the technique of Lagrange Multipliers, leads to four simultaneous equations which are quite tedious to solve esp. in an exam-setup under limited duration of time.
Entirely alternate solutions and/or techniques that efficiently solves the above system of equations are welcome.
multivariable-calculus lagrange-multiplier maxima-minima
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
edited Jan 13 at 18:49
Winged Blades of Godric
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
asked Jan 13 at 18:35
Winged Blades of GodricWinged Blades of Godric
635
635
1
$begingroup$
So it is $f(x, y, color{red}{z})$? or what is $z$?
$endgroup$
– caverac
Jan 13 at 18:47
$begingroup$
My fault; apologies:( Editing in a moment.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:48
$begingroup$
Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
$begingroup$
Is the constraint $g(x,y,z)=0$, then?
$endgroup$
– amd
Jan 13 at 20:44
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55
|
show 2 more comments
1
$begingroup$
So it is $f(x, y, color{red}{z})$? or what is $z$?
$endgroup$
– caverac
Jan 13 at 18:47
$begingroup$
My fault; apologies:( Editing in a moment.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:48
$begingroup$
Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
$begingroup$
Is the constraint $g(x,y,z)=0$, then?
$endgroup$
– amd
Jan 13 at 20:44
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55
1
1
$begingroup$
So it is $f(x, y, color{red}{z})$? or what is $z$?
$endgroup$
– caverac
Jan 13 at 18:47
$begingroup$
So it is $f(x, y, color{red}{z})$? or what is $z$?
$endgroup$
– caverac
Jan 13 at 18:47
$begingroup$
My fault; apologies:( Editing in a moment.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:48
$begingroup$
My fault; apologies:( Editing in a moment.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:48
$begingroup$
Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
$begingroup$
Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
$begingroup$
Is the constraint $g(x,y,z)=0$, then?
$endgroup$
– amd
Jan 13 at 20:44
$begingroup$
Is the constraint $g(x,y,z)=0$, then?
$endgroup$
– amd
Jan 13 at 20:44
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55
|
show 2 more comments
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1
$begingroup$
So it is $f(x, y, color{red}{z})$? or what is $z$?
$endgroup$
– caverac
Jan 13 at 18:47
$begingroup$
My fault; apologies:( Editing in a moment.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:48
$begingroup$
Unless you have more constraints, it looks like you will have to go through the tedious way. e.g. Are you looking for a maximum? or just an extremum in general?
$endgroup$
– caverac
Jan 13 at 19:07
$begingroup$
Is the constraint $g(x,y,z)=0$, then?
$endgroup$
– amd
Jan 13 at 20:44
$begingroup$
@amd Obviously :-)
$endgroup$
– Winged Blades of Godric
Jan 14 at 3:55