Minimum $M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$ for $3x^2+8y^3=20$












0












$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







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  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55
















0












$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55














0












0








0


1



$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







share|cite|improve this question











$endgroup$





Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima




calculus inequality optimization nonlinear-optimization maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:27









Martin Sleziak

44.8k10119272




44.8k10119272










asked Dec 13 '18 at 11:26









Trong TuanTrong Tuan

1318




1318








  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55














  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55








1




1




$begingroup$
What does > mean in the second line?
$endgroup$
– William Elliot
Dec 13 '18 at 11:55




$begingroup$
What does > mean in the second line?
$endgroup$
– William Elliot
Dec 13 '18 at 11:55










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $x=2$ and $y=1$.



Thus, we'll get a value $6$.



We'll prove that it's a minimal value.



Indeed, let $x=2a$ and $y=b$.



Thus, $3a^2+2b^3=5$ and we need to prove that
$$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
But by AM-GM
$$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
$$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
which gives $2geq a^2+b^2.$



Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
Now, let $a=tb$.



Thus, we need to prove that
$$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
$$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



But for $0<tleq1$ we obtain:
$$16t^4+16t^3-7t^2-2t+1=$$
$$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
Done!



Also, we can use derivative here.



Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
$$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
$$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28










  • $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50










  • $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38





















0












$begingroup$

Local minimum:
$$f(2,1)=6$$
Global minimum:
$$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
enter image description here






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38


















    3












    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38
















    3












    3








    3





    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$



    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 13:49

























    answered Dec 13 '18 at 12:04









    Michael RozenbergMichael Rozenberg

    106k1893198




    106k1893198












    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38




















    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38


















    $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28




    $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28












    $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50




    $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50












    $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38






    $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38













    0












    $begingroup$

    Local minimum:
    $$f(2,1)=6$$
    Global minimum:
    $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Local minimum:
      $$f(2,1)=6$$
      Global minimum:
      $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Local minimum:
        $$f(2,1)=6$$
        Global minimum:
        $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
        enter image description here






        share|cite|improve this answer









        $endgroup$



        Local minimum:
        $$f(2,1)=6$$
        Global minimum:
        $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 20:43









        Aleksas DomarkasAleksas Domarkas

        1,37216




        1,37216






























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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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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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