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







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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















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







share|cite|improve this question
























  • 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













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







share|cite|improve this question















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






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edited Nov 22 at 7:11









Jean-Claude Arbaut

14.7k63363




14.7k63363










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


















  • 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










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)$$






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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 $lambda_2=-2$).






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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      active

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      up vote
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      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)$$






      share|cite|improve this answer

























        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)$$






        share|cite|improve this answer























          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)$$






          share|cite|improve this answer












          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)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 7:35









          Mostafa Ayaz

          13.5k3836




          13.5k3836






















              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 $lambda_2=-2$).






              share|cite|improve this answer



























                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 $lambda_2=-2$).






                share|cite|improve this answer

























                  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 $lambda_2=-2$).






                  share|cite|improve this answer














                  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 $lambda_2=-2$).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 22 at 8:42

























                  answered Nov 22 at 7:28









                  José Carlos Santos

                  146k22116215




                  146k22116215






























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