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What is the minimum for this function of $x_1,x_2, \ldots, x_n$:

$$\sum_{i=1}^n c_i \log x_i + \lambda \; \sum_{i=1}^n d_i x_i, $$ where $\lambda$, $c$ and $d$ series are positive constants, $x_i \in (0.02,1]$ and $\sum x_i = 1$.

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how about letting any $x_i \to 0$, then your function goes to $-\infty$, so it does not have a minimum (assuming that the corresponding $c_i > 0$. – user1709 Nov 6 '10 at 15:57
Forgot to add.. $x_i \epsilon (0,1]$ – SkypeMeSM Nov 6 '10 at 16:02
A minimum, many minima. – Mariano Suárez-Alvarez Nov 6 '10 at 16:04
Another way of saying Slowsolver's comment: you have a ceiling, but certainly no floor. – J. M. Nov 6 '10 at 16:08
Right. its a concave function and so the minima is $-\infty$. I am thinking about adding some bounds and constraints. Something like $x_i \epsilon (0.02,1]$ and $\sum x_i = 1$. – SkypeMeSM Nov 6 '10 at 16:34

Let $ x_i=1/m $ certainly $ x_i \in (0,1] $ for all $i$ and all natural numbers $m$. Then your function becomes $-\log m \sum_{i=1}^{n}{c_i} + \frac{\lambda}{m} (\sum_{i=1}^{n}{d_i})$ which can be written as $-m A + B/m $ where $A,B$ are positive constants. This can be made arbitrarily large and negative, so there is no minimum. If this is part of some work on Lagrangian multipliers you need to think again about your restrictions on $\lambda$.

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Actually $\lambda$ is just a weight constant. Will the above bounds make any difference? – SkypeMeSM Nov 6 '10 at 16:51
Firstly what is the nature of this problem (i.e. why can you change the bounds?) Secondly using an open interval for your bounds on the xi means that any minimum is unlikely to be attained. – Charlie Nov 6 '10 at 16:59
The larger problem I am trying to solve is… . I am fine with assuming the bounds on xi, since the minimum is going to be constrained by the bounds instead of going to $-infty$. The tricky part though is getting the minimum while making sure that $\sum x_i = 1$, otherwise I can take all $x_i$ to be 0.02 and that will result in minimum value for this function. I hope I am not just talking nonsense. Please correct me, if I am. Thanks. – SkypeMeSM Nov 7 '10 at 15:25
I stated this question in a slightly different way here – SkypeMeSM Nov 7 '10 at 15:39

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