4
$\begingroup$

How to maximize this? $$ \sum\limits_{i=1}^n u_iln(x_i), $$ where $u_i,x_i$ are real numbers, $n$ is a positive integer, $0 \leq u_i \leq 1, 0 < x_i < 1, \sum\limits_{i=1}^n u_i = 1, \sum\limits_{i=1}^n x_i = 1$, and only $x_i$ are variables, others are given numbers.

$\endgroup$
3
$\begingroup$

Hint: Use Lagrange multipliers: http://en.wikipedia.org/wiki/Lagrange_multiplier

In this case, you have to work with

$\Lambda(x_1,...,x_n,\lambda)=\sum u_i \log(x_i)-\lambda(x_1+...+x_n-1)$

and follow the method described in Wikipedia.

In fact we have

$\dfrac{\partial \Lambda}{\partial x_i}=\dfrac{u_i}{x_i}-\lambda$ for all $i$, so all the $\dfrac{u_i}{x_i}$ are equal (to $\lambda$). So the solution is to take $x_i=u_i$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If there was no condition for the $u_i$, the solution would correspond to $x_i=\frac{u_i}{\sum_{i=1}^n u_i}$ $\endgroup$ – Claude Leibovici Apr 22 '15 at 8:16
  • $\begingroup$ @ClaudeLeibovici Exactly! $\endgroup$ – mathifold.org Apr 22 '15 at 8:22
  • $\begingroup$ @mathifold.org But you can't take $x_i = u_i$ if $u_i = 0$. So what's the problem using Lagrange multipliers? $\endgroup$ – Daniel Apr 22 '15 at 8:32
  • $\begingroup$ @thuzhf If $u_i=0$, the term does not longer exist, so taking $x_i=0$ should not be a problem, even when we take its logarithm. If $a=u_i=x_i$, then $\lim_{a\longrightarrow 0}a\log(a)=0$ $\endgroup$ – mathifold.org Apr 22 '15 at 8:38
  • $\begingroup$ @mathifold.org But if $u_i = 0$, how can you solve $\frac {\partial \Lambda} {\partial x_i} = \frac {u_i} {x_i} - \lambda = 0$, and finally get $x_i = u_i$? $\endgroup$ – Daniel Apr 22 '15 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.