Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$












2














Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$



My try:



By Lagrange Multiplier method we have



$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$



For $$L_x=0$$ we get



$$2x+lambda+4mu x=0 tag{1}$$



For $$L_y=0$$ we get



$$12y+2lambda+2mu y=0 tag{2}$$



For $$L_z=0$$ we get



$$8z+lambda=0 tag{3}$$



From $(1)$ and $(2)$ we get



$$x=frac{4 lambda}{1-2mu}$$



$$y=frac{8 lambda}{6-mu}$$



Substituting $x$ , $y$ and $z$ above in constrainst we get



$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$



$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$



But its tedious to solve above equations for $lambda$ and $mu$



Any other approach?






optimization systems-of-equations lagrange-multiplier maxima-minima







share|cite|improve this question






















  • WLOG $sqrt2x=4cos t,y=4sin t, z=?$
    – lab bhattacharjee
    Nov 22 at 13:43










  • Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
    – Ekaveera Kumar Sharma
    Nov 22 at 13:58












  • Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
    – Alex Silva
    Nov 22 at 15:29


















2














Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$



My try:



By Lagrange Multiplier method we have



$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$



For $$L_x=0$$ we get



$$2x+lambda+4mu x=0 tag{1}$$



For $$L_y=0$$ we get



$$12y+2lambda+2mu y=0 tag{2}$$



For $$L_z=0$$ we get



$$8z+lambda=0 tag{3}$$



From $(1)$ and $(2)$ we get



$$x=frac{4 lambda}{1-2mu}$$



$$y=frac{8 lambda}{6-mu}$$



Substituting $x$ , $y$ and $z$ above in constrainst we get



$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$



$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$



But its tedious to solve above equations for $lambda$ and $mu$



Any other approach?






optimization systems-of-equations lagrange-multiplier maxima-minima







share|cite|improve this question






















  • WLOG $sqrt2x=4cos t,y=4sin t, z=?$
    – lab bhattacharjee
    Nov 22 at 13:43










  • Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
    – Ekaveera Kumar Sharma
    Nov 22 at 13:58












  • Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
    – Alex Silva
    Nov 22 at 15:29
















2












2








2


2





Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$



My try:



By Lagrange Multiplier method we have



$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$



For $$L_x=0$$ we get



$$2x+lambda+4mu x=0 tag{1}$$



For $$L_y=0$$ we get



$$12y+2lambda+2mu y=0 tag{2}$$



For $$L_z=0$$ we get



$$8z+lambda=0 tag{3}$$



From $(1)$ and $(2)$ we get



$$x=frac{4 lambda}{1-2mu}$$



$$y=frac{8 lambda}{6-mu}$$



Substituting $x$ , $y$ and $z$ above in constrainst we get



$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$



$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$



But its tedious to solve above equations for $lambda$ and $mu$



Any other approach?






optimization systems-of-equations lagrange-multiplier maxima-minima







share|cite|improve this question













Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$



My try:



By Lagrange Multiplier method we have



$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$



For $$L_x=0$$ we get



$$2x+lambda+4mu x=0 tag{1}$$



For $$L_y=0$$ we get



$$12y+2lambda+2mu y=0 tag{2}$$



For $$L_z=0$$ we get



$$8z+lambda=0 tag{3}$$



From $(1)$ and $(2)$ we get



$$x=frac{4 lambda}{1-2mu}$$



$$y=frac{8 lambda}{6-mu}$$



Substituting $x$ , $y$ and $z$ above in constrainst we get



$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$



$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$



But its tedious to solve above equations for $lambda$ and $mu$



Any other approach?






optimization systems-of-equations lagrange-multiplier maxima-minima




optimization systems-of-equations lagrange-multiplier maxima-minima






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 13:37









Ekaveera Kumar Sharma

5,53511328




5,53511328












  • WLOG $sqrt2x=4cos t,y=4sin t, z=?$
    – lab bhattacharjee
    Nov 22 at 13:43










  • Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
    – Ekaveera Kumar Sharma
    Nov 22 at 13:58












  • Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
    – Alex Silva
    Nov 22 at 15:29




















  • WLOG $sqrt2x=4cos t,y=4sin t, z=?$
    – lab bhattacharjee
    Nov 22 at 13:43










  • Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
    – Ekaveera Kumar Sharma
    Nov 22 at 13:58












  • Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
    – Alex Silva
    Nov 22 at 15:29


















WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43




WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43












Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58






Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58














Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29






Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29












3 Answers
3






active

oldest

votes


















1














Hint:



Let $u=x^2+6y^2+4(4-2y-x)^2$



$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$



As $x$ is real, the discriminant will be $ge0$



$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$



$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$



$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$



Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$






share|cite|improve this answer





















  • Very nice approach
    – Ekaveera Kumar Sharma
    Nov 23 at 7:10



















1














Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.






share|cite|improve this answer























  • Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
    – Ekaveera Kumar Sharma
    Nov 22 at 14:29



















0














With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Hint:



    Let $u=x^2+6y^2+4(4-2y-x)^2$



    $iff5x^2-16(2-y)x+22y^2-16y+16-u=0$



    As $x$ is real, the discriminant will be $ge0$



    $$implies256(y-2)^2ge20(22y^2-16y+16-u)$$



    $$implies64(y-2)^2ge5(22y^2-16y+16-u)$$



    $$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$



    Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$






    share|cite|improve this answer





















    • Very nice approach
      – Ekaveera Kumar Sharma
      Nov 23 at 7:10
















    1














    Hint:



    Let $u=x^2+6y^2+4(4-2y-x)^2$



    $iff5x^2-16(2-y)x+22y^2-16y+16-u=0$



    As $x$ is real, the discriminant will be $ge0$



    $$implies256(y-2)^2ge20(22y^2-16y+16-u)$$



    $$implies64(y-2)^2ge5(22y^2-16y+16-u)$$



    $$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$



    Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$






    share|cite|improve this answer





















    • Very nice approach
      – Ekaveera Kumar Sharma
      Nov 23 at 7:10














    1












    1








    1






    Hint:



    Let $u=x^2+6y^2+4(4-2y-x)^2$



    $iff5x^2-16(2-y)x+22y^2-16y+16-u=0$



    As $x$ is real, the discriminant will be $ge0$



    $$implies256(y-2)^2ge20(22y^2-16y+16-u)$$



    $$implies64(y-2)^2ge5(22y^2-16y+16-u)$$



    $$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$



    Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$






    share|cite|improve this answer












    Hint:



    Let $u=x^2+6y^2+4(4-2y-x)^2$



    $iff5x^2-16(2-y)x+22y^2-16y+16-u=0$



    As $x$ is real, the discriminant will be $ge0$



    $$implies256(y-2)^2ge20(22y^2-16y+16-u)$$



    $$implies64(y-2)^2ge5(22y^2-16y+16-u)$$



    $$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$



    Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 23 at 6:26









    lab bhattacharjee

    222k15155273




    222k15155273












    • Very nice approach
      – Ekaveera Kumar Sharma
      Nov 23 at 7:10


















    • Very nice approach
      – Ekaveera Kumar Sharma
      Nov 23 at 7:10
















    Very nice approach
    – Ekaveera Kumar Sharma
    Nov 23 at 7:10




    Very nice approach
    – Ekaveera Kumar Sharma
    Nov 23 at 7:10











    1














    Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.






    share|cite|improve this answer























    • Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
      – Ekaveera Kumar Sharma
      Nov 22 at 14:29
















    1














    Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.






    share|cite|improve this answer























    • Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
      – Ekaveera Kumar Sharma
      Nov 22 at 14:29














    1












    1








    1






    Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.






    share|cite|improve this answer














    Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 22 at 14:21

























    answered Nov 22 at 13:59









    José Carlos Santos

    148k22117218




    148k22117218












    • Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
      – Ekaveera Kumar Sharma
      Nov 22 at 14:29


















    • Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
      – Ekaveera Kumar Sharma
      Nov 22 at 14:29
















    Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
    – Ekaveera Kumar Sharma
    Nov 22 at 14:29




    Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
    – Ekaveera Kumar Sharma
    Nov 22 at 14:29











    0














    With computer I get numeric solution
    $$f_{min}=9.445967377367024$$
    if
    $$x=2.811441051354938\
    y=0.4377195786259503\
    z=0.3131197913931612$$
    Exact values of $x,y,z$ is solutions of equations:
    $$2033x^4-800x^3-13960x^2-1792x+6144=0,\
    2033y^4-6016y^3-27920y^2+88064y-32768=0,\
    2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
    We can solve these equations exact and get exact solutions.
    However, there are too big expressions.






    share|cite|improve this answer


























      0














      With computer I get numeric solution
      $$f_{min}=9.445967377367024$$
      if
      $$x=2.811441051354938\
      y=0.4377195786259503\
      z=0.3131197913931612$$
      Exact values of $x,y,z$ is solutions of equations:
      $$2033x^4-800x^3-13960x^2-1792x+6144=0,\
      2033y^4-6016y^3-27920y^2+88064y-32768=0,\
      2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
      We can solve these equations exact and get exact solutions.
      However, there are too big expressions.






      share|cite|improve this answer
























        0












        0








        0






        With computer I get numeric solution
        $$f_{min}=9.445967377367024$$
        if
        $$x=2.811441051354938\
        y=0.4377195786259503\
        z=0.3131197913931612$$
        Exact values of $x,y,z$ is solutions of equations:
        $$2033x^4-800x^3-13960x^2-1792x+6144=0,\
        2033y^4-6016y^3-27920y^2+88064y-32768=0,\
        2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
        We can solve these equations exact and get exact solutions.
        However, there are too big expressions.






        share|cite|improve this answer












        With computer I get numeric solution
        $$f_{min}=9.445967377367024$$
        if
        $$x=2.811441051354938\
        y=0.4377195786259503\
        z=0.3131197913931612$$
        Exact values of $x,y,z$ is solutions of equations:
        $$2033x^4-800x^3-13960x^2-1792x+6144=0,\
        2033y^4-6016y^3-27920y^2+88064y-32768=0,\
        2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
        We can solve these equations exact and get exact solutions.
        However, there are too big expressions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 11:23









        Aleksas Domarkas

        8276




        8276






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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