I need to prove the maxima of the following summation, using Lagrange. max x m ( ∑ m a m l o g ( x m ) ) s.t. 0 ≤ x m ≤ 1 ∑ m x m = 1The solution is a closed form x m = a m ∑ m a m .I formulated the Lagrange equation but I am confused about the signs and the multipliers. L ( x , λ , μ ) = ∑ m a m l o g ( x m ) + ∑ m λ m ( 1 − x m ) + μ ( ∑ m x m − 1 ), Is this formulation correct ? what is wrong ?note: only one μ for one constraint.

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I need to prove the maxima of the following summation, using Lagrange.      max                  x        m                  (          ∑      m              a      m        l    o    g    (          x      m        )    )  s.t.  0  ≤      x    m    ≤  1      ∑    m        x    m    =  1The solution is a closed form       x    m    =            a      m                      ∑        m                    a        m             .I formulated the Lagrange equation but I am confused about the signs and the multipliers.  L  (  x  ,  λ  ,  μ  )  =      ∑    m        a    m    l  o  g  (      x    m    )  +      ∑    m        λ    m    (  1  −      x    m    )  +  μ  (      ∑    m        x    m    −  1  ), Is this formulation correct ? what is wrong ?note: only one   μ for one constraint.

Timothy Mcclure · Accepted Answer

I think that the Lagrangian is:  L  (  x  ,  λ  ,  μ  )  =      ∑    m        a    m    l  o  g  (      x    m    )  +  μ      (          ∑      m              x      m        −    1    )  Now we have:      {                                                      a              1                                      x              1                                +          μ          =          0                                                                a              2                                      x              2                                +          μ          =          0                                                                                    a              m                                      x              m                                +          μ          =          0                                                  ∑            m                                x            m                    =          1                        so      {                                        x            1                    =          −                                    a              1                        μ                                                            x            2                    =          −                                    a              2                        μ                                                                                x            m                    =          −                                    a              m                        μ                                                            ∑            m                                x            m                    =          1                        and from the last equation:   −      1    μ        ∑    m        a    m    =  1 so   μ  =  −      ∑    m        a    m  . We can write the system like this:      {                                        x            1                    =          −                                    a              1                                      −                              ∑                m                                            a                m                                                                                      x            2                    =          −                                    a              2                                      −                              ∑                m                                            a                m                                                                                                          x            m                    =          −                                    a              m                                      −                              ∑                m                                            a                m                                                                          μ          =          −                      ∑            m                                a            m                                        {                                        x            1                    =                                    a              1                                                      ∑                m                                            a                m                                                                                      x            2                    =                                    a              2                                                      ∑                m                                            a                m                                                                                                          x            m                    =                                    a              m                                                      ∑                m                                            a                m                                                                          μ          =          −                      ∑            m                                a            m

I need to prove the maxima of the following summation, using Lagrange. <munder> <mo mova

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