A Problem of Lagrange Multiplier
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The problem is find the minimum value of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$.
Let $f(x,y,z)=x^2+y^2+z^2$, then after some calculation I got this two equations:
$4+6lambda_1+lambda_2(1-f)=0 $ and $2f+lambda_1-3lambda_2=0$
Now I can solve these two equations to find $lambda_1$ and $lambda_2$ in terms for $f$. Now I cant understand how to proceed.
Please help with explanation.
multivariable-calculus lagrange-multiplier maxima-minima
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 7:10
nonei
111
add a comment |
up vote
2
down vote
favorite
The problem is find the minimum value of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$.
Let $f(x,y,z)=x^2+y^2+z^2$, then after some calculation I got this two equations:
$4+6lambda_1+lambda_2(1-f)=0 $ and $2f+lambda_1-3lambda_2=0$
Now I can solve these two equations to find $lambda_1$ and $lambda_2$ in terms for $f$. Now I cant understand how to proceed.
Please help with explanation.
multivariable-calculus lagrange-multiplier maxima-minima
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 7:10
nonei
111
You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The problem is find the minimum value of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$.
Let $f(x,y,z)=x^2+y^2+z^2$, then after some calculation I got this two equations:
$4+6lambda_1+lambda_2(1-f)=0 $ and $2f+lambda_1-3lambda_2=0$
Now I can solve these two equations to find $lambda_1$ and $lambda_2$ in terms for $f$. Now I cant understand how to proceed.
Please help with explanation.
multivariable-calculus lagrange-multiplier maxima-minima
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 7:10
nonei
111
The problem is find the minimum value of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$.
Let $f(x,y,z)=x^2+y^2+z^2$, then after some calculation I got this two equations:
$4+6lambda_1+lambda_2(1-f)=0 $ and $2f+lambda_1-3lambda_2=0$
Now I can solve these two equations to find $lambda_1$ and $lambda_2$ in terms for $f$. Now I cant understand how to proceed.
Please help with explanation.
multivariable-calculus lagrange-multiplier maxima-minima
multivariable-calculus lagrange-multiplier maxima-minima
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 7:10
nonei
111
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 7:10
nonei
111
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
edited Nov 22 at 7:11
Jean-Claude Arbaut
14.7k63363
14.7k63363
asked Nov 22 at 7:10
nonei
111
asked Nov 22 at 7:10
nonei
111
asked Nov 22 at 7:10
nonei
111
111
You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28
add a comment |
You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28
You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28
add a comment |
2 Answers
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Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$or equivalently$$2x^2=lambda_1 x-lambda_2\2y^2=lambda_1 y-lambda_2\2z^2=lambda_1 z-lambda_2$$the equation $2u^2-lambda_1 u+lambda_2=0$ has two roots as following$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=yne z$
in this case $$2x+z=1\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $lambda_1=0\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\(1,-1,1)\(-1,1,1)$$
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
add a comment |
up vote
1
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I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$left{begin{array}{l}2x=lambda_1+lambda_2yz\2y=lambda_1+lambda_2xz\2z=lambda_1+lambda_2xy\x+y+z=1\xyz=-1.end{array}right.$$There are only $3$ ponts in $mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $lambda_1=0$ and $lambda_2=-2$).
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$or equivalently$$2x^2=lambda_1 x-lambda_2\2y^2=lambda_1 y-lambda_2\2z^2=lambda_1 z-lambda_2$$the equation $2u^2-lambda_1 u+lambda_2=0$ has two roots as following$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=yne z$
in this case $$2x+z=1\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $lambda_1=0\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\(1,-1,1)\(-1,1,1)$$
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
add a comment |
up vote
1
down vote
Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$or equivalently$$2x^2=lambda_1 x-lambda_2\2y^2=lambda_1 y-lambda_2\2z^2=lambda_1 z-lambda_2$$the equation $2u^2-lambda_1 u+lambda_2=0$ has two roots as following$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=yne z$
in this case $$2x+z=1\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $lambda_1=0\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\(1,-1,1)\(-1,1,1)$$
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
add a comment |
up vote
1
down vote
up vote
1
down vote
Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$or equivalently$$2x^2=lambda_1 x-lambda_2\2y^2=lambda_1 y-lambda_2\2z^2=lambda_1 z-lambda_2$$the equation $2u^2-lambda_1 u+lambda_2=0$ has two roots as following$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=yne z$
in this case $$2x+z=1\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $lambda_1=0\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\(1,-1,1)\(-1,1,1)$$
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$or equivalently$$2x^2=lambda_1 x-lambda_2\2y^2=lambda_1 y-lambda_2\2z^2=lambda_1 z-lambda_2$$the equation $2u^2-lambda_1 u+lambda_2=0$ has two roots as following$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=yne z$
in this case $$2x+z=1\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $lambda_1=0\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\(1,-1,1)\(-1,1,1)$$
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
answered Nov 22 at 7:35
Mostafa Ayaz
13.5k3836
13.5k3836
add a comment |
add a comment |
up vote
1
down vote
I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$left{begin{array}{l}2x=lambda_1+lambda_2yz\2y=lambda_1+lambda_2xz\2z=lambda_1+lambda_2xy\x+y+z=1\xyz=-1.end{array}right.$$There are only $3$ ponts in $mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $lambda_1=0$ and $lambda_2=-2$).
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
add a comment |
up vote
1
down vote
I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$left{begin{array}{l}2x=lambda_1+lambda_2yz\2y=lambda_1+lambda_2xz\2z=lambda_1+lambda_2xy\x+y+z=1\xyz=-1.end{array}right.$$There are only $3$ ponts in $mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $lambda_1=0$ and $lambda_2=-2$).
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$left{begin{array}{l}2x=lambda_1+lambda_2yz\2y=lambda_1+lambda_2xz\2z=lambda_1+lambda_2xy\x+y+z=1\xyz=-1.end{array}right.$$There are only $3$ ponts in $mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $lambda_1=0$ and $lambda_2=-2$).
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$left{begin{array}{l}2x=lambda_1+lambda_2yz\2y=lambda_1+lambda_2xz\2z=lambda_1+lambda_2xy\x+y+z=1\xyz=-1.end{array}right.$$There are only $3$ ponts in $mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $lambda_1=0$ and $lambda_2=-2$).
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
edited Nov 22 at 8:42
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
answered Nov 22 at 7:28
José Carlos Santos
146k22116215
146k22116215
add a comment |
add a comment |
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You have two equations in two unknowns, $lambda_1$ and $lambda_2$. Solve!
– David G. Stork
Nov 22 at 7:18
Why is $f$ in your equations?
– smcc
Nov 22 at 7:22
can anybody come with a complete solution...i solved for that two unknown multiplier....then what?
– nonei
Nov 22 at 7:28