How can I get exact maximal value of this expression?Finding maximum value with position from Table of valuesHow do I find the position of the maximum value in each column of a table?Finding the associated parameter of the maximumHow to maximize the modulus of a multivariate complex-valued function?Finding maximum in the boundary limits using mathematicaNSum: Summand (or its derivative) is not numerical at pointHow to do Maximum Likelihood Estimation with this Multivariate Normal?NMaximize is not converging to a solutionUnable to find maximum value of my custom functionParametric Find Maximum

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How can I get exact maximal value of this expression?


Finding maximum value with position from Table of valuesHow do I find the position of the maximum value in each column of a table?Finding the associated parameter of the maximumHow to maximize the modulus of a multivariate complex-valued function?Finding maximum in the boundary limits using mathematicaNSum: Summand (or its derivative) is not numerical at pointHow to do Maximum Likelihood Estimation with this Multivariate Normal?NMaximize is not converging to a solutionUnable to find maximum value of my custom functionParametric Find Maximum













3












$begingroup$


I am trying to find the exact maximum value of the expression
$$ E= sqrt5 a^2+a (4 b-2 c)+2b^2+4 b c+5 c^2+sqrt2a^2+a (2 b+2 c)+2 b^2-2 bc+2 c^2+sqrt26 a^2+a (10c-2 b)+26 b^2+10 b c+2 c^2 ,$$ where $a^2 + b^2 + c^2 = 1$ and $a, b, c > 0$.



I know that, the answer is $ 4sqrt3 + sqrt6 .$
When I tried



NMaximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


I got the approximate answer



9.37769, a -> 0.801784, b -> 0.534523, c -> 0.267261


When I tried,



Maximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


It's take about 3 minutes, I could't get the answer. How can I get exact maximize value of that expression?










share|improve this question











$endgroup$
















    3












    $begingroup$


    I am trying to find the exact maximum value of the expression
    $$ E= sqrt5 a^2+a (4 b-2 c)+2b^2+4 b c+5 c^2+sqrt2a^2+a (2 b+2 c)+2 b^2-2 bc+2 c^2+sqrt26 a^2+a (10c-2 b)+26 b^2+10 b c+2 c^2 ,$$ where $a^2 + b^2 + c^2 = 1$ and $a, b, c > 0$.



    I know that, the answer is $ 4sqrt3 + sqrt6 .$
    When I tried



    NMaximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
    Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
    Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
    a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


    I got the approximate answer



    9.37769, a -> 0.801784, b -> 0.534523, c -> 0.267261


    When I tried,



    Maximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
    Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
    Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
    a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


    It's take about 3 minutes, I could't get the answer. How can I get exact maximize value of that expression?










    share|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I am trying to find the exact maximum value of the expression
      $$ E= sqrt5 a^2+a (4 b-2 c)+2b^2+4 b c+5 c^2+sqrt2a^2+a (2 b+2 c)+2 b^2-2 bc+2 c^2+sqrt26 a^2+a (10c-2 b)+26 b^2+10 b c+2 c^2 ,$$ where $a^2 + b^2 + c^2 = 1$ and $a, b, c > 0$.



      I know that, the answer is $ 4sqrt3 + sqrt6 .$
      When I tried



      NMaximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
      Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
      Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
      a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


      I got the approximate answer



      9.37769, a -> 0.801784, b -> 0.534523, c -> 0.267261


      When I tried,



      Maximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
      Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
      Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
      a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


      It's take about 3 minutes, I could't get the answer. How can I get exact maximize value of that expression?










      share|improve this question











      $endgroup$




      I am trying to find the exact maximum value of the expression
      $$ E= sqrt5 a^2+a (4 b-2 c)+2b^2+4 b c+5 c^2+sqrt2a^2+a (2 b+2 c)+2 b^2-2 bc+2 c^2+sqrt26 a^2+a (10c-2 b)+26 b^2+10 b c+2 c^2 ,$$ where $a^2 + b^2 + c^2 = 1$ and $a, b, c > 0$.



      I know that, the answer is $ 4sqrt3 + sqrt6 .$
      When I tried



      NMaximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
      Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
      Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
      a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


      I got the approximate answer



      9.37769, a -> 0.801784, b -> 0.534523, c -> 0.267261


      When I tried,



      Maximize[(Sqrt[2*a^2 + (2*b + 2*c)*a + 2*b^2 - 2*b*c + 2*c^2] + 
      Sqrt[5*a^2 + (4*b - 2*c)*a + 2*b^2 + 4*b*c + 5*c^2] +
      Sqrt[26*a^2 + (-2*b + 10*c)*a + 26*b^2 + 10*b*c + 2*c^2]),
      a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0, a, b, c]


      It's take about 3 minutes, I could't get the answer. How can I get exact maximize value of that expression?







      maximum






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 27 at 4:08







      minhthien_2016

















      asked May 26 at 12:58









      minhthien_2016minhthien_2016

      600311




      600311




















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          Writing:



          rad1 = Sqrt[5 a^2 + 4 a b + 2 b^2 - 2 a c + 4 b c + 5 c^2];
          rad2 = Sqrt[2] Sqrt[a^2 + b^2 - b c + c^2 + a (b + c)];
          rad3 = Sqrt[2] Sqrt[13 a^2 - a b + 13 b^2 + 5 (a + b) c + c^2];
          str = "PossibleClosedForm", 1, "FormulaData";

          fct = rad1 + rad2 + rad3;
          bond = a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0;
          sol1 = NMaximize[fct, bond, a, b, c, WorkingPrecision -> 40];

          max = WolframAlpha[ToString[sol1[[1]]], str][[1, 1]];
          a0 = WolframAlpha[ToString[sol1[[2, 1, 2]]], str][[1, 1]];
          b0 = WolframAlpha[ToString[sol1[[2, 2, 2]]], str][[1, 1]];
          c0 = WolframAlpha[ToString[sol1[[2, 3, 2]]], str][[1, 1]];

          sol2 = max, a -> a0, b -> b0, c -> c0
          sol1 == N[sol2]


          I get:




          4 Sqrt[3] + Sqrt[6], a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14]



          True




          which is what is desired.






          share|improve this answer











          $endgroup$




















            7












            $begingroup$

            I don't know why Maximize is unable to find the maximum. Instead of using Maximize, you could try using Lagrange multipliers. To make the algebra easier, define:



            constraint1 = 2*a^2+(2*b+2*c)*a+2*b^2-2*b*c+2*c^2 - d^2;
            constraint2 = 5*a^2+(4*b-2*c)*a+2*b^2+4*b*c+5*c^2 - e^2;
            constraint3 = 26*a^2+(-2*b+10*c)*a+26*b^2+10*b*c+2*c^2 - f^2;
            constraint4 = a^2 + b^2 + c^2 - 1;


            Then, the goal is to maximize d + e + f subject to having each constraint equal to 0 (I'm ignoring the positivity constraint for now). Using Lagrange multipliers, we want to extremize the function:



            obj = d + e + f + λ1 constraint1 + λ2 constraint2 + λ3 constraint3 + λ4 constraint4;


            Setting the derivatives with respect to each of the unknowns to 0:



            eqns = Thread[D[obj, a, b, c, d, e, f, λ1, λ2, λ3, λ4] == 0];


            Let's solve them:



            sols = Solve[eqns, a, b, c, d, e, f, λ1, λ2, λ3, λ4, Reals];//AbsoluteTiming



            1.77698, Null




            Now, let's impose the constraints that a, b, c, d, e, and f are all positive:



            r = Cases[sols, x_ /; AllTrue[a, b, c, d, e, f /. x, GreaterEqualThan[0]]]



            a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14], d -> Sqrt[3],
            e -> Sqrt[6],
            f -> 3 Sqrt[3], λ1 -> 1/(2 Sqrt[3]), λ2 -> 1/(
            2 Sqrt[6]), λ3 -> 1/(6 Sqrt[3]), λ4 ->
            1/2 (-4 Sqrt[3] - Sqrt[6])




            So, the maximum value is:



            d + e + f /. First [r]



            4 Sqrt[3] + Sqrt[6]







            share|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









              7












              $begingroup$

              Writing:



              rad1 = Sqrt[5 a^2 + 4 a b + 2 b^2 - 2 a c + 4 b c + 5 c^2];
              rad2 = Sqrt[2] Sqrt[a^2 + b^2 - b c + c^2 + a (b + c)];
              rad3 = Sqrt[2] Sqrt[13 a^2 - a b + 13 b^2 + 5 (a + b) c + c^2];
              str = "PossibleClosedForm", 1, "FormulaData";

              fct = rad1 + rad2 + rad3;
              bond = a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0;
              sol1 = NMaximize[fct, bond, a, b, c, WorkingPrecision -> 40];

              max = WolframAlpha[ToString[sol1[[1]]], str][[1, 1]];
              a0 = WolframAlpha[ToString[sol1[[2, 1, 2]]], str][[1, 1]];
              b0 = WolframAlpha[ToString[sol1[[2, 2, 2]]], str][[1, 1]];
              c0 = WolframAlpha[ToString[sol1[[2, 3, 2]]], str][[1, 1]];

              sol2 = max, a -> a0, b -> b0, c -> c0
              sol1 == N[sol2]


              I get:




              4 Sqrt[3] + Sqrt[6], a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14]



              True




              which is what is desired.






              share|improve this answer











              $endgroup$

















                7












                $begingroup$

                Writing:



                rad1 = Sqrt[5 a^2 + 4 a b + 2 b^2 - 2 a c + 4 b c + 5 c^2];
                rad2 = Sqrt[2] Sqrt[a^2 + b^2 - b c + c^2 + a (b + c)];
                rad3 = Sqrt[2] Sqrt[13 a^2 - a b + 13 b^2 + 5 (a + b) c + c^2];
                str = "PossibleClosedForm", 1, "FormulaData";

                fct = rad1 + rad2 + rad3;
                bond = a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0;
                sol1 = NMaximize[fct, bond, a, b, c, WorkingPrecision -> 40];

                max = WolframAlpha[ToString[sol1[[1]]], str][[1, 1]];
                a0 = WolframAlpha[ToString[sol1[[2, 1, 2]]], str][[1, 1]];
                b0 = WolframAlpha[ToString[sol1[[2, 2, 2]]], str][[1, 1]];
                c0 = WolframAlpha[ToString[sol1[[2, 3, 2]]], str][[1, 1]];

                sol2 = max, a -> a0, b -> b0, c -> c0
                sol1 == N[sol2]


                I get:




                4 Sqrt[3] + Sqrt[6], a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14]



                True




                which is what is desired.






                share|improve this answer











                $endgroup$















                  7












                  7








                  7





                  $begingroup$

                  Writing:



                  rad1 = Sqrt[5 a^2 + 4 a b + 2 b^2 - 2 a c + 4 b c + 5 c^2];
                  rad2 = Sqrt[2] Sqrt[a^2 + b^2 - b c + c^2 + a (b + c)];
                  rad3 = Sqrt[2] Sqrt[13 a^2 - a b + 13 b^2 + 5 (a + b) c + c^2];
                  str = "PossibleClosedForm", 1, "FormulaData";

                  fct = rad1 + rad2 + rad3;
                  bond = a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0;
                  sol1 = NMaximize[fct, bond, a, b, c, WorkingPrecision -> 40];

                  max = WolframAlpha[ToString[sol1[[1]]], str][[1, 1]];
                  a0 = WolframAlpha[ToString[sol1[[2, 1, 2]]], str][[1, 1]];
                  b0 = WolframAlpha[ToString[sol1[[2, 2, 2]]], str][[1, 1]];
                  c0 = WolframAlpha[ToString[sol1[[2, 3, 2]]], str][[1, 1]];

                  sol2 = max, a -> a0, b -> b0, c -> c0
                  sol1 == N[sol2]


                  I get:




                  4 Sqrt[3] + Sqrt[6], a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14]



                  True




                  which is what is desired.






                  share|improve this answer











                  $endgroup$



                  Writing:



                  rad1 = Sqrt[5 a^2 + 4 a b + 2 b^2 - 2 a c + 4 b c + 5 c^2];
                  rad2 = Sqrt[2] Sqrt[a^2 + b^2 - b c + c^2 + a (b + c)];
                  rad3 = Sqrt[2] Sqrt[13 a^2 - a b + 13 b^2 + 5 (a + b) c + c^2];
                  str = "PossibleClosedForm", 1, "FormulaData";

                  fct = rad1 + rad2 + rad3;
                  bond = a^2 + b^2 + c^2 == 1, a > 0, b > 0, c > 0;
                  sol1 = NMaximize[fct, bond, a, b, c, WorkingPrecision -> 40];

                  max = WolframAlpha[ToString[sol1[[1]]], str][[1, 1]];
                  a0 = WolframAlpha[ToString[sol1[[2, 1, 2]]], str][[1, 1]];
                  b0 = WolframAlpha[ToString[sol1[[2, 2, 2]]], str][[1, 1]];
                  c0 = WolframAlpha[ToString[sol1[[2, 3, 2]]], str][[1, 1]];

                  sol2 = max, a -> a0, b -> b0, c -> c0
                  sol1 == N[sol2]


                  I get:




                  4 Sqrt[3] + Sqrt[6], a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14]



                  True




                  which is what is desired.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited May 26 at 17:11

























                  answered May 26 at 17:05









                  TeMTeM

                  2,131621




                  2,131621





















                      7












                      $begingroup$

                      I don't know why Maximize is unable to find the maximum. Instead of using Maximize, you could try using Lagrange multipliers. To make the algebra easier, define:



                      constraint1 = 2*a^2+(2*b+2*c)*a+2*b^2-2*b*c+2*c^2 - d^2;
                      constraint2 = 5*a^2+(4*b-2*c)*a+2*b^2+4*b*c+5*c^2 - e^2;
                      constraint3 = 26*a^2+(-2*b+10*c)*a+26*b^2+10*b*c+2*c^2 - f^2;
                      constraint4 = a^2 + b^2 + c^2 - 1;


                      Then, the goal is to maximize d + e + f subject to having each constraint equal to 0 (I'm ignoring the positivity constraint for now). Using Lagrange multipliers, we want to extremize the function:



                      obj = d + e + f + λ1 constraint1 + λ2 constraint2 + λ3 constraint3 + λ4 constraint4;


                      Setting the derivatives with respect to each of the unknowns to 0:



                      eqns = Thread[D[obj, a, b, c, d, e, f, λ1, λ2, λ3, λ4] == 0];


                      Let's solve them:



                      sols = Solve[eqns, a, b, c, d, e, f, λ1, λ2, λ3, λ4, Reals];//AbsoluteTiming



                      1.77698, Null




                      Now, let's impose the constraints that a, b, c, d, e, and f are all positive:



                      r = Cases[sols, x_ /; AllTrue[a, b, c, d, e, f /. x, GreaterEqualThan[0]]]



                      a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14], d -> Sqrt[3],
                      e -> Sqrt[6],
                      f -> 3 Sqrt[3], λ1 -> 1/(2 Sqrt[3]), λ2 -> 1/(
                      2 Sqrt[6]), λ3 -> 1/(6 Sqrt[3]), λ4 ->
                      1/2 (-4 Sqrt[3] - Sqrt[6])




                      So, the maximum value is:



                      d + e + f /. First [r]



                      4 Sqrt[3] + Sqrt[6]







                      share|improve this answer









                      $endgroup$

















                        7












                        $begingroup$

                        I don't know why Maximize is unable to find the maximum. Instead of using Maximize, you could try using Lagrange multipliers. To make the algebra easier, define:



                        constraint1 = 2*a^2+(2*b+2*c)*a+2*b^2-2*b*c+2*c^2 - d^2;
                        constraint2 = 5*a^2+(4*b-2*c)*a+2*b^2+4*b*c+5*c^2 - e^2;
                        constraint3 = 26*a^2+(-2*b+10*c)*a+26*b^2+10*b*c+2*c^2 - f^2;
                        constraint4 = a^2 + b^2 + c^2 - 1;


                        Then, the goal is to maximize d + e + f subject to having each constraint equal to 0 (I'm ignoring the positivity constraint for now). Using Lagrange multipliers, we want to extremize the function:



                        obj = d + e + f + λ1 constraint1 + λ2 constraint2 + λ3 constraint3 + λ4 constraint4;


                        Setting the derivatives with respect to each of the unknowns to 0:



                        eqns = Thread[D[obj, a, b, c, d, e, f, λ1, λ2, λ3, λ4] == 0];


                        Let's solve them:



                        sols = Solve[eqns, a, b, c, d, e, f, λ1, λ2, λ3, λ4, Reals];//AbsoluteTiming



                        1.77698, Null




                        Now, let's impose the constraints that a, b, c, d, e, and f are all positive:



                        r = Cases[sols, x_ /; AllTrue[a, b, c, d, e, f /. x, GreaterEqualThan[0]]]



                        a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14], d -> Sqrt[3],
                        e -> Sqrt[6],
                        f -> 3 Sqrt[3], λ1 -> 1/(2 Sqrt[3]), λ2 -> 1/(
                        2 Sqrt[6]), λ3 -> 1/(6 Sqrt[3]), λ4 ->
                        1/2 (-4 Sqrt[3] - Sqrt[6])




                        So, the maximum value is:



                        d + e + f /. First [r]



                        4 Sqrt[3] + Sqrt[6]







                        share|improve this answer









                        $endgroup$















                          7












                          7








                          7





                          $begingroup$

                          I don't know why Maximize is unable to find the maximum. Instead of using Maximize, you could try using Lagrange multipliers. To make the algebra easier, define:



                          constraint1 = 2*a^2+(2*b+2*c)*a+2*b^2-2*b*c+2*c^2 - d^2;
                          constraint2 = 5*a^2+(4*b-2*c)*a+2*b^2+4*b*c+5*c^2 - e^2;
                          constraint3 = 26*a^2+(-2*b+10*c)*a+26*b^2+10*b*c+2*c^2 - f^2;
                          constraint4 = a^2 + b^2 + c^2 - 1;


                          Then, the goal is to maximize d + e + f subject to having each constraint equal to 0 (I'm ignoring the positivity constraint for now). Using Lagrange multipliers, we want to extremize the function:



                          obj = d + e + f + λ1 constraint1 + λ2 constraint2 + λ3 constraint3 + λ4 constraint4;


                          Setting the derivatives with respect to each of the unknowns to 0:



                          eqns = Thread[D[obj, a, b, c, d, e, f, λ1, λ2, λ3, λ4] == 0];


                          Let's solve them:



                          sols = Solve[eqns, a, b, c, d, e, f, λ1, λ2, λ3, λ4, Reals];//AbsoluteTiming



                          1.77698, Null




                          Now, let's impose the constraints that a, b, c, d, e, and f are all positive:



                          r = Cases[sols, x_ /; AllTrue[a, b, c, d, e, f /. x, GreaterEqualThan[0]]]



                          a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14], d -> Sqrt[3],
                          e -> Sqrt[6],
                          f -> 3 Sqrt[3], λ1 -> 1/(2 Sqrt[3]), λ2 -> 1/(
                          2 Sqrt[6]), λ3 -> 1/(6 Sqrt[3]), λ4 ->
                          1/2 (-4 Sqrt[3] - Sqrt[6])




                          So, the maximum value is:



                          d + e + f /. First [r]



                          4 Sqrt[3] + Sqrt[6]







                          share|improve this answer









                          $endgroup$



                          I don't know why Maximize is unable to find the maximum. Instead of using Maximize, you could try using Lagrange multipliers. To make the algebra easier, define:



                          constraint1 = 2*a^2+(2*b+2*c)*a+2*b^2-2*b*c+2*c^2 - d^2;
                          constraint2 = 5*a^2+(4*b-2*c)*a+2*b^2+4*b*c+5*c^2 - e^2;
                          constraint3 = 26*a^2+(-2*b+10*c)*a+26*b^2+10*b*c+2*c^2 - f^2;
                          constraint4 = a^2 + b^2 + c^2 - 1;


                          Then, the goal is to maximize d + e + f subject to having each constraint equal to 0 (I'm ignoring the positivity constraint for now). Using Lagrange multipliers, we want to extremize the function:



                          obj = d + e + f + λ1 constraint1 + λ2 constraint2 + λ3 constraint3 + λ4 constraint4;


                          Setting the derivatives with respect to each of the unknowns to 0:



                          eqns = Thread[D[obj, a, b, c, d, e, f, λ1, λ2, λ3, λ4] == 0];


                          Let's solve them:



                          sols = Solve[eqns, a, b, c, d, e, f, λ1, λ2, λ3, λ4, Reals];//AbsoluteTiming



                          1.77698, Null




                          Now, let's impose the constraints that a, b, c, d, e, and f are all positive:



                          r = Cases[sols, x_ /; AllTrue[a, b, c, d, e, f /. x, GreaterEqualThan[0]]]



                          a -> 3/Sqrt[14], b -> Sqrt[2/7], c -> 1/Sqrt[14], d -> Sqrt[3],
                          e -> Sqrt[6],
                          f -> 3 Sqrt[3], λ1 -> 1/(2 Sqrt[3]), λ2 -> 1/(
                          2 Sqrt[6]), λ3 -> 1/(6 Sqrt[3]), λ4 ->
                          1/2 (-4 Sqrt[3] - Sqrt[6])




                          So, the maximum value is:



                          d + e + f /. First [r]



                          4 Sqrt[3] + Sqrt[6]








                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered May 26 at 20:33









                          Carl WollCarl Woll

                          83.2k3105216




                          83.2k3105216



























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