maximum value of algebraic expression (another)












-1












$begingroup$


if $p^2+q^2+r^2=5$ and $p,q,r$ all are real number,



then maximum value of $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



what i try . Expanding $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



$41p^2+41q^2+25r^2-24pq-40qr-30pr$



$25times 5+16p^2+16q^2-24pq-40qr-30pr$



How i use inequality to find maximum of given expression



Help me please






inequality optimization maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Lagrange_multiplier
    $endgroup$
    – Federico
    Dec 17 '18 at 19:38










  • $begingroup$
    The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
    $endgroup$
    – Federico
    Dec 17 '18 at 19:40






  • 1




    $begingroup$
    you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
    $endgroup$
    – Will Jagy
    Dec 17 '18 at 20:01






  • 1




    $begingroup$
    Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
    $endgroup$
    – Jyrki Lahtonen
    Dec 17 '18 at 20:35
















-1












$begingroup$


if $p^2+q^2+r^2=5$ and $p,q,r$ all are real number,



then maximum value of $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



what i try . Expanding $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



$41p^2+41q^2+25r^2-24pq-40qr-30pr$



$25times 5+16p^2+16q^2-24pq-40qr-30pr$



How i use inequality to find maximum of given expression



Help me please






inequality optimization maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Lagrange_multiplier
    $endgroup$
    – Federico
    Dec 17 '18 at 19:38










  • $begingroup$
    The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
    $endgroup$
    – Federico
    Dec 17 '18 at 19:40






  • 1




    $begingroup$
    you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
    $endgroup$
    – Will Jagy
    Dec 17 '18 at 20:01






  • 1




    $begingroup$
    Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
    $endgroup$
    – Jyrki Lahtonen
    Dec 17 '18 at 20:35














-1












-1








-1





$begingroup$


if $p^2+q^2+r^2=5$ and $p,q,r$ all are real number,



then maximum value of $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



what i try . Expanding $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



$41p^2+41q^2+25r^2-24pq-40qr-30pr$



$25times 5+16p^2+16q^2-24pq-40qr-30pr$



How i use inequality to find maximum of given expression



Help me please






inequality optimization maxima-minima







share|cite|improve this question











$endgroup$




if $p^2+q^2+r^2=5$ and $p,q,r$ all are real number,



then maximum value of $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



what i try . Expanding $(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$



$41p^2+41q^2+25r^2-24pq-40qr-30pr$



$25times 5+16p^2+16q^2-24pq-40qr-30pr$



How i use inequality to find maximum of given expression



Help me please






inequality optimization maxima-minima




inequality 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 17 '18 at 21:13









Michael Rozenberg

104k1891196




104k1891196










asked Dec 17 '18 at 19:36









jackyjacky

861612




861612








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Lagrange_multiplier
    $endgroup$
    – Federico
    Dec 17 '18 at 19:38










  • $begingroup$
    The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
    $endgroup$
    – Federico
    Dec 17 '18 at 19:40






  • 1




    $begingroup$
    you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
    $endgroup$
    – Will Jagy
    Dec 17 '18 at 20:01






  • 1




    $begingroup$
    Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
    $endgroup$
    – Jyrki Lahtonen
    Dec 17 '18 at 20:35














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Lagrange_multiplier
    $endgroup$
    – Federico
    Dec 17 '18 at 19:38










  • $begingroup$
    The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
    $endgroup$
    – Federico
    Dec 17 '18 at 19:40






  • 1




    $begingroup$
    you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
    $endgroup$
    – Will Jagy
    Dec 17 '18 at 20:01






  • 1




    $begingroup$
    Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
    $endgroup$
    – Jyrki Lahtonen
    Dec 17 '18 at 20:35








1




1




$begingroup$
en.wikipedia.org/wiki/Lagrange_multiplier
$endgroup$
– Federico
Dec 17 '18 at 19:38




$begingroup$
en.wikipedia.org/wiki/Lagrange_multiplier
$endgroup$
– Federico
Dec 17 '18 at 19:38












$begingroup$
The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
$endgroup$
– Federico
Dec 17 '18 at 19:40




$begingroup$
The maximum is $250$, achieved at $p=-frac{4}{sqrt{5}}, q =frac{3}{sqrt{5}}, r=0$.
$endgroup$
– Federico
Dec 17 '18 at 19:40




1




1




$begingroup$
you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
$endgroup$
– Will Jagy
Dec 17 '18 at 20:01




$begingroup$
you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$
$endgroup$
– Will Jagy
Dec 17 '18 at 20:01




1




1




$begingroup$
Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 20:35




$begingroup$
Please check out our guide for new askers. This is very much missing in context. Are you expected to just use some trickery (high school level tools extended for contests), did this come up in a multivariable calculus course when we could expect you to be familiar with Lagrange multipliers? Et cetera.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 20:35










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$begingroup$

We'll prove that $250$ it's a maximal value.



Indeed,
$$250geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's
$$50(p^2+q^2+r^2)geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or
$$(3p+4q+5r)^2geq0.$$
The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$






share|cite|improve this answer











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    $begingroup$

    We'll prove that $250$ it's a maximal value.



    Indeed,
    $$250geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's
    $$50(p^2+q^2+r^2)geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or
    $$(3p+4q+5r)^2geq0.$$
    The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      We'll prove that $250$ it's a maximal value.



      Indeed,
      $$250geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's
      $$50(p^2+q^2+r^2)geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or
      $$(3p+4q+5r)^2geq0.$$
      The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        We'll prove that $250$ it's a maximal value.



        Indeed,
        $$250geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's
        $$50(p^2+q^2+r^2)geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or
        $$(3p+4q+5r)^2geq0.$$
        The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$






        share|cite|improve this answer











        $endgroup$



        We'll prove that $250$ it's a maximal value.



        Indeed,
        $$250geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's
        $$50(p^2+q^2+r^2)geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or
        $$(3p+4q+5r)^2geq0.$$
        The equality occurs for $p^2+q^2+r^2=5$ and $3p+4q+5r=0.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 20:50

























        answered Dec 17 '18 at 20:42









        Michael RozenbergMichael Rozenberg

        104k1891196




        104k1891196






























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