Find maximum value under constraints by Lagrange multipliers or another method












0












$begingroup$


Let $f(x,y)=(x + 2y)^2 + (3x + 4y)^2.$



Then for $(x,y)$ on the circle $x^2+y^2=1 $ what is the maximum value of $f$?



I tried to use Lagrange multiplier method with the constraint $g(x,y)= x^2 + y^2 =1.$



However, now I realize that this way is so complicated on this problem
($lambda = 15 pm sqrt{221}$)



How can I solve this problem?






lagrange-multiplier







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  • $begingroup$
    Did you try using polar coordinates?
    $endgroup$
    – Damien
    Jan 11 at 6:09
















0












$begingroup$


Let $f(x,y)=(x + 2y)^2 + (3x + 4y)^2.$



Then for $(x,y)$ on the circle $x^2+y^2=1 $ what is the maximum value of $f$?



I tried to use Lagrange multiplier method with the constraint $g(x,y)= x^2 + y^2 =1.$



However, now I realize that this way is so complicated on this problem
($lambda = 15 pm sqrt{221}$)



How can I solve this problem?






lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you try using polar coordinates?
    $endgroup$
    – Damien
    Jan 11 at 6:09














0












0








0





$begingroup$


Let $f(x,y)=(x + 2y)^2 + (3x + 4y)^2.$



Then for $(x,y)$ on the circle $x^2+y^2=1 $ what is the maximum value of $f$?



I tried to use Lagrange multiplier method with the constraint $g(x,y)= x^2 + y^2 =1.$



However, now I realize that this way is so complicated on this problem
($lambda = 15 pm sqrt{221}$)



How can I solve this problem?






lagrange-multiplier







share|cite|improve this question











$endgroup$




Let $f(x,y)=(x + 2y)^2 + (3x + 4y)^2.$



Then for $(x,y)$ on the circle $x^2+y^2=1 $ what is the maximum value of $f$?



I tried to use Lagrange multiplier method with the constraint $g(x,y)= x^2 + y^2 =1.$



However, now I realize that this way is so complicated on this problem
($lambda = 15 pm sqrt{221}$)



How can I solve this problem?






lagrange-multiplier




lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 6:14









David G. Stork

10.7k31332




10.7k31332










asked Jan 11 at 5:55









hewhew

306




306












  • $begingroup$
    Did you try using polar coordinates?
    $endgroup$
    – Damien
    Jan 11 at 6:09


















  • $begingroup$
    Did you try using polar coordinates?
    $endgroup$
    – Damien
    Jan 11 at 6:09
















$begingroup$
Did you try using polar coordinates?
$endgroup$
– Damien
Jan 11 at 6:09




$begingroup$
Did you try using polar coordinates?
$endgroup$
– Damien
Jan 11 at 6:09










5 Answers
5






active

oldest

votes


















2












$begingroup$

If Lagrange's is not mandatory, WLOG choose $x=cos t, y=sin t$ to find



$$f(x,y)=Acos^2t+Bsin tcos t+Csin^2t=g(t)text{(say)}$$



Use $cos2t=1-2sin^2t=2cos^2t-1,sin2t=2sin tcos t$ to find $g(t)$ to be of the form $$pcos2t+qsin2t+r$$



Now use $-sqrt{p^2+q^2}le pcos2t+qsin2tlesqrt{p^2+q^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!!
    $endgroup$
    – hew
    Jan 11 at 6:35



















1












$begingroup$

This is simple enough that you can use your constraint as a substitution, i.e., $y = sqrt{1 - x^2}$.



Then your function is:



$$f(x) = left(2 sqrt{1-x^2}+xright)^2+left(4 sqrt{1-x^2}+3 xright)^2$$



enter image description here



Its derivative is:



$${d f(x) over dx} = -frac{4 left(14 x^2+5 sqrt{1-x^2} x-7right)}{sqrt{1-x^2}}$$



Set it to zero and find:



$$x = left{-sqrt{frac{1}{2}+frac{5}{2 sqrt{221}}},sqrt{frac{1}{442} left(221-5 sqrt{221}right)}right} approx { -0.817416, 0.576048 }$$



where clearly one is a minimum, the other a maximum.



For "culture," here is the figure in three dimensions:



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much!!
    $endgroup$
    – hew
    Jan 11 at 6:34










  • $begingroup$
    @hew: So perhaps a (+1) is in order?
    $endgroup$
    – David G. Stork
    Jan 11 at 6:36



















0












$begingroup$

Exact solution with Lagrange multiplier method is:
$$f_{min}=15-sqrt{221}approx0.134$$
if
$$x=frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx0.817,;
y=-frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx-0.576$$
or
$$x=-frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx-0.817,;
y=frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx0.576$$
enter image description here






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is possible to find the solution by using the polar coordinates:



    $$f(theta) = 10cos^2theta + 20sin^2theta + 28sinthetacostheta = 10 + 10sin^2theta + 14sin2theta$$



    The extrema are found by deriving:



    $$f'(theta) = 20sintheta costheta + 28cos2theta = 10sin2theta + 28cos2theta = 0$$



    And we get
    $$theta = -frac{1}{2} arctanfrac{14}{5} + kfrac{pi}{2} $$
    where $k$ is any integer between $0$ and $3$



    We can check numerically that the maximum corresponds to
    $$theta = -frac{1}{2} arctanfrac{14}{5} + kpi approx -0.6138 + kpi$$



    where $k$ is equal to $0$ or $1$



    This corresponds to $x approx 0.81754, ,y approx -0.576$ and $x approx -0.81754, ,y approx 0.576$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $lambda^2-30lambda+4$.






      share|cite|improve this answer









      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        If Lagrange's is not mandatory, WLOG choose $x=cos t, y=sin t$ to find



        $$f(x,y)=Acos^2t+Bsin tcos t+Csin^2t=g(t)text{(say)}$$



        Use $cos2t=1-2sin^2t=2cos^2t-1,sin2t=2sin tcos t$ to find $g(t)$ to be of the form $$pcos2t+qsin2t+r$$



        Now use $-sqrt{p^2+q^2}le pcos2t+qsin2tlesqrt{p^2+q^2}$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:35
















        2












        $begingroup$

        If Lagrange's is not mandatory, WLOG choose $x=cos t, y=sin t$ to find



        $$f(x,y)=Acos^2t+Bsin tcos t+Csin^2t=g(t)text{(say)}$$



        Use $cos2t=1-2sin^2t=2cos^2t-1,sin2t=2sin tcos t$ to find $g(t)$ to be of the form $$pcos2t+qsin2t+r$$



        Now use $-sqrt{p^2+q^2}le pcos2t+qsin2tlesqrt{p^2+q^2}$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:35














        2












        2








        2





        $begingroup$

        If Lagrange's is not mandatory, WLOG choose $x=cos t, y=sin t$ to find



        $$f(x,y)=Acos^2t+Bsin tcos t+Csin^2t=g(t)text{(say)}$$



        Use $cos2t=1-2sin^2t=2cos^2t-1,sin2t=2sin tcos t$ to find $g(t)$ to be of the form $$pcos2t+qsin2t+r$$



        Now use $-sqrt{p^2+q^2}le pcos2t+qsin2tlesqrt{p^2+q^2}$






        share|cite|improve this answer









        $endgroup$



        If Lagrange's is not mandatory, WLOG choose $x=cos t, y=sin t$ to find



        $$f(x,y)=Acos^2t+Bsin tcos t+Csin^2t=g(t)text{(say)}$$



        Use $cos2t=1-2sin^2t=2cos^2t-1,sin2t=2sin tcos t$ to find $g(t)$ to be of the form $$pcos2t+qsin2t+r$$



        Now use $-sqrt{p^2+q^2}le pcos2t+qsin2tlesqrt{p^2+q^2}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 6:08









        lab bhattacharjeelab bhattacharjee

        224k15156274




        224k15156274












        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:35


















        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:35
















        $begingroup$
        Thank you very much!!
        $endgroup$
        – hew
        Jan 11 at 6:35




        $begingroup$
        Thank you very much!!
        $endgroup$
        – hew
        Jan 11 at 6:35











        1












        $begingroup$

        This is simple enough that you can use your constraint as a substitution, i.e., $y = sqrt{1 - x^2}$.



        Then your function is:



        $$f(x) = left(2 sqrt{1-x^2}+xright)^2+left(4 sqrt{1-x^2}+3 xright)^2$$



        enter image description here



        Its derivative is:



        $${d f(x) over dx} = -frac{4 left(14 x^2+5 sqrt{1-x^2} x-7right)}{sqrt{1-x^2}}$$



        Set it to zero and find:



        $$x = left{-sqrt{frac{1}{2}+frac{5}{2 sqrt{221}}},sqrt{frac{1}{442} left(221-5 sqrt{221}right)}right} approx { -0.817416, 0.576048 }$$



        where clearly one is a minimum, the other a maximum.



        For "culture," here is the figure in three dimensions:



        enter image description here






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:34










        • $begingroup$
          @hew: So perhaps a (+1) is in order?
          $endgroup$
          – David G. Stork
          Jan 11 at 6:36
















        1












        $begingroup$

        This is simple enough that you can use your constraint as a substitution, i.e., $y = sqrt{1 - x^2}$.



        Then your function is:



        $$f(x) = left(2 sqrt{1-x^2}+xright)^2+left(4 sqrt{1-x^2}+3 xright)^2$$



        enter image description here



        Its derivative is:



        $${d f(x) over dx} = -frac{4 left(14 x^2+5 sqrt{1-x^2} x-7right)}{sqrt{1-x^2}}$$



        Set it to zero and find:



        $$x = left{-sqrt{frac{1}{2}+frac{5}{2 sqrt{221}}},sqrt{frac{1}{442} left(221-5 sqrt{221}right)}right} approx { -0.817416, 0.576048 }$$



        where clearly one is a minimum, the other a maximum.



        For "culture," here is the figure in three dimensions:



        enter image description here






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:34










        • $begingroup$
          @hew: So perhaps a (+1) is in order?
          $endgroup$
          – David G. Stork
          Jan 11 at 6:36














        1












        1








        1





        $begingroup$

        This is simple enough that you can use your constraint as a substitution, i.e., $y = sqrt{1 - x^2}$.



        Then your function is:



        $$f(x) = left(2 sqrt{1-x^2}+xright)^2+left(4 sqrt{1-x^2}+3 xright)^2$$



        enter image description here



        Its derivative is:



        $${d f(x) over dx} = -frac{4 left(14 x^2+5 sqrt{1-x^2} x-7right)}{sqrt{1-x^2}}$$



        Set it to zero and find:



        $$x = left{-sqrt{frac{1}{2}+frac{5}{2 sqrt{221}}},sqrt{frac{1}{442} left(221-5 sqrt{221}right)}right} approx { -0.817416, 0.576048 }$$



        where clearly one is a minimum, the other a maximum.



        For "culture," here is the figure in three dimensions:



        enter image description here






        share|cite|improve this answer











        $endgroup$



        This is simple enough that you can use your constraint as a substitution, i.e., $y = sqrt{1 - x^2}$.



        Then your function is:



        $$f(x) = left(2 sqrt{1-x^2}+xright)^2+left(4 sqrt{1-x^2}+3 xright)^2$$



        enter image description here



        Its derivative is:



        $${d f(x) over dx} = -frac{4 left(14 x^2+5 sqrt{1-x^2} x-7right)}{sqrt{1-x^2}}$$



        Set it to zero and find:



        $$x = left{-sqrt{frac{1}{2}+frac{5}{2 sqrt{221}}},sqrt{frac{1}{442} left(221-5 sqrt{221}right)}right} approx { -0.817416, 0.576048 }$$



        where clearly one is a minimum, the other a maximum.



        For "culture," here is the figure in three dimensions:



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 11 at 19:19

























        answered Jan 11 at 6:05









        David G. StorkDavid G. Stork

        10.7k31332




        10.7k31332












        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:34










        • $begingroup$
          @hew: So perhaps a (+1) is in order?
          $endgroup$
          – David G. Stork
          Jan 11 at 6:36


















        • $begingroup$
          Thank you very much!!
          $endgroup$
          – hew
          Jan 11 at 6:34










        • $begingroup$
          @hew: So perhaps a (+1) is in order?
          $endgroup$
          – David G. Stork
          Jan 11 at 6:36
















        $begingroup$
        Thank you very much!!
        $endgroup$
        – hew
        Jan 11 at 6:34




        $begingroup$
        Thank you very much!!
        $endgroup$
        – hew
        Jan 11 at 6:34












        $begingroup$
        @hew: So perhaps a (+1) is in order?
        $endgroup$
        – David G. Stork
        Jan 11 at 6:36




        $begingroup$
        @hew: So perhaps a (+1) is in order?
        $endgroup$
        – David G. Stork
        Jan 11 at 6:36











        0












        $begingroup$

        Exact solution with Lagrange multiplier method is:
        $$f_{min}=15-sqrt{221}approx0.134$$
        if
        $$x=frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx0.817,;
        y=-frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx-0.576$$
        or
        $$x=-frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx-0.817,;
        y=frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx0.576$$
        enter image description here






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Exact solution with Lagrange multiplier method is:
          $$f_{min}=15-sqrt{221}approx0.134$$
          if
          $$x=frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx0.817,;
          y=-frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx-0.576$$
          or
          $$x=-frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx-0.817,;
          y=frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx0.576$$
          enter image description here






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Exact solution with Lagrange multiplier method is:
            $$f_{min}=15-sqrt{221}approx0.134$$
            if
            $$x=frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx0.817,;
            y=-frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx-0.576$$
            or
            $$x=-frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx-0.817,;
            y=frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx0.576$$
            enter image description here






            share|cite|improve this answer









            $endgroup$



            Exact solution with Lagrange multiplier method is:
            $$f_{min}=15-sqrt{221}approx0.134$$
            if
            $$x=frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx0.817,;
            y=-frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx-0.576$$
            or
            $$x=-frac{sqrt{5 sqrt{221}+221}}{sqrt{442}}approx-0.817,;
            y=frac{sqrt{221-5 sqrt{221}}}{sqrt{442}}approx0.576$$
            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 11 at 8:11









            Aleksas DomarkasAleksas Domarkas

            8976




            8976























                0












                $begingroup$

                It is possible to find the solution by using the polar coordinates:



                $$f(theta) = 10cos^2theta + 20sin^2theta + 28sinthetacostheta = 10 + 10sin^2theta + 14sin2theta$$



                The extrema are found by deriving:



                $$f'(theta) = 20sintheta costheta + 28cos2theta = 10sin2theta + 28cos2theta = 0$$



                And we get
                $$theta = -frac{1}{2} arctanfrac{14}{5} + kfrac{pi}{2} $$
                where $k$ is any integer between $0$ and $3$



                We can check numerically that the maximum corresponds to
                $$theta = -frac{1}{2} arctanfrac{14}{5} + kpi approx -0.6138 + kpi$$



                where $k$ is equal to $0$ or $1$



                This corresponds to $x approx 0.81754, ,y approx -0.576$ and $x approx -0.81754, ,y approx 0.576$






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  It is possible to find the solution by using the polar coordinates:



                  $$f(theta) = 10cos^2theta + 20sin^2theta + 28sinthetacostheta = 10 + 10sin^2theta + 14sin2theta$$



                  The extrema are found by deriving:



                  $$f'(theta) = 20sintheta costheta + 28cos2theta = 10sin2theta + 28cos2theta = 0$$



                  And we get
                  $$theta = -frac{1}{2} arctanfrac{14}{5} + kfrac{pi}{2} $$
                  where $k$ is any integer between $0$ and $3$



                  We can check numerically that the maximum corresponds to
                  $$theta = -frac{1}{2} arctanfrac{14}{5} + kpi approx -0.6138 + kpi$$



                  where $k$ is equal to $0$ or $1$



                  This corresponds to $x approx 0.81754, ,y approx -0.576$ and $x approx -0.81754, ,y approx 0.576$






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    It is possible to find the solution by using the polar coordinates:



                    $$f(theta) = 10cos^2theta + 20sin^2theta + 28sinthetacostheta = 10 + 10sin^2theta + 14sin2theta$$



                    The extrema are found by deriving:



                    $$f'(theta) = 20sintheta costheta + 28cos2theta = 10sin2theta + 28cos2theta = 0$$



                    And we get
                    $$theta = -frac{1}{2} arctanfrac{14}{5} + kfrac{pi}{2} $$
                    where $k$ is any integer between $0$ and $3$



                    We can check numerically that the maximum corresponds to
                    $$theta = -frac{1}{2} arctanfrac{14}{5} + kpi approx -0.6138 + kpi$$



                    where $k$ is equal to $0$ or $1$



                    This corresponds to $x approx 0.81754, ,y approx -0.576$ and $x approx -0.81754, ,y approx 0.576$






                    share|cite|improve this answer











                    $endgroup$



                    It is possible to find the solution by using the polar coordinates:



                    $$f(theta) = 10cos^2theta + 20sin^2theta + 28sinthetacostheta = 10 + 10sin^2theta + 14sin2theta$$



                    The extrema are found by deriving:



                    $$f'(theta) = 20sintheta costheta + 28cos2theta = 10sin2theta + 28cos2theta = 0$$



                    And we get
                    $$theta = -frac{1}{2} arctanfrac{14}{5} + kfrac{pi}{2} $$
                    where $k$ is any integer between $0$ and $3$



                    We can check numerically that the maximum corresponds to
                    $$theta = -frac{1}{2} arctanfrac{14}{5} + kpi approx -0.6138 + kpi$$



                    where $k$ is equal to $0$ or $1$



                    This corresponds to $x approx 0.81754, ,y approx -0.576$ and $x approx -0.81754, ,y approx 0.576$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 11 at 9:37

























                    answered Jan 11 at 9:18









                    DamienDamien

                    60714




                    60714























                        0












                        $begingroup$

                        The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $lambda^2-30lambda+4$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $lambda^2-30lambda+4$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $lambda^2-30lambda+4$.






                            share|cite|improve this answer









                            $endgroup$



                            The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $lambda^2-30lambda+4$.







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                            answered Jan 11 at 9:53









                            amdamd

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