How to maximize $\sum\limits_{i=1}^n u_iln(x_i)$?

Question

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.

Answer 1

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 $x_i=u_i$.