Maximizing a weighted sum of logarithms











up vote
1
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I'm trying to show that for:



$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$



Where:



$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$



That:



$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$



I'm really not certain where even to begin with this derivation.






inequality optimization summation logarithms convex-optimization







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  • Try the KKT approach.
    – LinAlg
    Nov 14 at 21:36










  • @LinAlg what's a good resource for KKT? I'm not super familiar.
    – mmam
    Nov 14 at 22:08










  • en.wikipedia.org/wiki/…
    – LinAlg
    Nov 14 at 23:02















up vote
1
down vote

favorite












I'm trying to show that for:



$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$



Where:



$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$



That:



$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$



I'm really not certain where even to begin with this derivation.






inequality optimization summation logarithms convex-optimization







share|cite|improve this question
























  • Try the KKT approach.
    – LinAlg
    Nov 14 at 21:36










  • @LinAlg what's a good resource for KKT? I'm not super familiar.
    – mmam
    Nov 14 at 22:08










  • en.wikipedia.org/wiki/…
    – LinAlg
    Nov 14 at 23:02













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to show that for:



$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$



Where:



$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$



That:



$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$



I'm really not certain where even to begin with this derivation.






inequality optimization summation logarithms convex-optimization







share|cite|improve this question















I'm trying to show that for:



$$p^* equiv max_{x} sum^{k=n}_{k=1} a_k ln x_k
$$



Where:



$
x in mathbb{R}^n \
x geq 0 \
sum_{k=1}^{k=n} x_k=c \
c > 0 \
forall a_k, a_k > 0 \
a equiv sum_{k=1}^{k=n} a_k
$



That:



$$
p^* = aln{frac{c}{a}}+sum^{k=n}_{k=1} a_k ln a_k
$$



I'm really not certain where even to begin with this derivation.






inequality optimization summation logarithms convex-optimization




inequality optimization summation logarithms convex-optimization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 17:57









Batominovski

33.5k33292




33.5k33292










asked Nov 14 at 20:59









mmam

811826




811826












  • Try the KKT approach.
    – LinAlg
    Nov 14 at 21:36










  • @LinAlg what's a good resource for KKT? I'm not super familiar.
    – mmam
    Nov 14 at 22:08










  • en.wikipedia.org/wiki/…
    – LinAlg
    Nov 14 at 23:02


















  • Try the KKT approach.
    – LinAlg
    Nov 14 at 21:36










  • @LinAlg what's a good resource for KKT? I'm not super familiar.
    – mmam
    Nov 14 at 22:08










  • en.wikipedia.org/wiki/…
    – LinAlg
    Nov 14 at 23:02
















Try the KKT approach.
– LinAlg
Nov 14 at 21:36




Try the KKT approach.
– LinAlg
Nov 14 at 21:36












@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08




@LinAlg what's a good resource for KKT? I'm not super familiar.
– mmam
Nov 14 at 22:08












en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02




en.wikipedia.org/wiki/…
– LinAlg
Nov 14 at 23:02










2 Answers
2






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I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.



We have that
$$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
with
$$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
By the Weighted AM-GM Inequality, we have
$$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
Take logarithm on both sides of the inequality above, we get
$$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
Ergo,
$$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
or equivalently,
$$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$






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    up vote
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    down vote













    With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.



    Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.



    Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.



    So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.



    So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$






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      2 Answers
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      2 Answers
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      up vote
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      I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.



      We have that
      $$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
      with
      $$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
      By the Weighted AM-GM Inequality, we have
      $$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
      Take logarithm on both sides of the inequality above, we get
      $$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
      Ergo,
      $$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
      Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
      or equivalently,
      $$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
      This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$






      share|cite|improve this answer

























        up vote
        2
        down vote













        I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.



        We have that
        $$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
        with
        $$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
        By the Weighted AM-GM Inequality, we have
        $$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
        Take logarithm on both sides of the inequality above, we get
        $$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
        Ergo,
        $$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
        Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
        or equivalently,
        $$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
        This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.



          We have that
          $$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
          with
          $$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
          By the Weighted AM-GM Inequality, we have
          $$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
          Take logarithm on both sides of the inequality above, we get
          $$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
          Ergo,
          $$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
          Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
          or equivalently,
          $$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
          This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$






          share|cite|improve this answer












          I shall come back with a Lagrangian approach tomorrow. However, I now present a solution using the Weighted AM-GM Inequality.



          We have that
          $$sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)=sum_{k=1}^n,x_k=c,,$$
          with
          $$sum_{k=1}^n,frac{a_i}{a}=frac{sumlimits_{k=1}^n,a_k}{a}=frac{a}{a}=1,.$$
          By the Weighted AM-GM Inequality, we have
          $$c=sum_{k=1}^n,frac{a_k}{a},left(frac{a,x_k}{a_k}right)geq prod_{k=1}^n,left(frac{a,x_k}{a_k}right)^{frac{a_k}{a}},.$$
          Take logarithm on both sides of the inequality above, we get
          $$ln(c)geq sum_{k=1}^n,frac{a_k}{a},lnleft(frac{a,x_k}{a_k}right)=frac{1}{a},sum_{k=1}^n,Big(a_k,lnleft(x_kright)-a_k,lnleft(a_kright)+a_k,ln(a)Big),.$$
          Ergo,
          $$sum_{k=1}^n,a_k,lnleft(x_kright)leq a,ln(c)-left(sum_{k=1}^n,a_kright),ln(a)+sum_{k=1}^n,a_k,lnleft(a_kright)=a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$
          Note that inequality above becomes an equality if and only if $$frac{x_1}{a_1}=frac{x_2}{a_2}=ldots=frac{x_n}{a_n}=frac{c}{a},,$$
          or equivalently,
          $$left(x_1,x_2,ldots,x_nright)=left(frac{c,a_1}{a},frac{c,a_2}{a},ldots,frac{c,a_n}{a}right),.$$
          This shows that $$p^*= a,lnleft(frac{c}{a}right)+sum_{k=1}^n,a_k,lnleft(a_kright),.$$







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          answered Nov 22 at 18:09









          Batominovski

          33.5k33292




          33.5k33292






















              up vote
              0
              down vote













              With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.



              Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.



              Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.



              So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.



              So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$






              share|cite|improve this answer

























                up vote
                0
                down vote













                With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.



                Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.



                Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.



                So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.



                So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$






                share|cite|improve this answer























                  up vote
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                  down vote










                  up vote
                  0
                  down vote









                  With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.



                  Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.



                  Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.



                  So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.



                  So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$






                  share|cite|improve this answer












                  With the Lagrange multiplier method you want to maximize $p(x)=sum a_k ln(x_k) +lambda (c-sum x_k)$.



                  Differentiating w.r.t. the $x$'s and setting equal to zero gives $frac{a_k}{x_k}=lambda$.



                  Imposing the constraint $c=sum x_k=frac{sum a_k}{lambda}=frac{a}{lambda}$. It follows that $lambda=a/c$.



                  So $x_k=frac{a_k }{ lambda}=frac{c a_k}{a}$.



                  So $p^*=sum a_k ln(x_k)=sum a_k ln(frac{c a_k}{a})=sum a_k (ln(frac{c}{a}) + ln(a_k))=aln(frac{c}{a})+sum a_k ln(a_k)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 18:37









                  user121049

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                  1,352164






























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