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

44.8k9118272




44.8k9118272










asked Dec 13 '18 at 11:26









Trong TuanTrong Tuan

1268




1268








  • 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

    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


















    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

    102k1791195




    102k1791195












    • $begingroup$
      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

        1,24916




        1,24916






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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