I am trying to solve the following problem:
Find the maximum with regards to each $x_i$ in the following objective function $$ 2\sum_i^N c_i d_i log(x_i) $$
subject to $k$ constraints: $$ \forall k: \sum_i^N c_i f_{ik} x_i = b_k $$
Of course we can write the constraints also as $Ax=b$ where $A$ is $k\times n$ matrix and $b$ is a vector of length $k$.
Here $c_i$ are just some constants that appear both in the objective function and in the constraints. And $d_i$ and $f_{ik}$ are some other constants.
I am not sure where to start at all. Lagrangian multipliers maybe?
What also holds in my case is that $\sum_k f_{ik}=1$ for any $i$. But I am guessing that fact is not necessary to solve the optimization. Also I can say that $b_k=1$ for example if it makes it easier, but again I was hoping for a general solution.
This is only a partial answer:
You can write down the Lagrangian. If $\lambda_j$ is the Lagrange multiplier associated with constraint $j\in\{1,\ldots,k\}$, then the first-order conditions are simply:
$$\frac{2c_id_i}{x_i}=\sum_{j=1}^k \lambda _jc_if_{ij}$$
for each $i\in\{1,\ldots,N\}$.
Thus
$$x_i=\frac{2d_i}{\sum_{j=1}^k \lambda_jf_{ij}}$$
If you substitute this into constraint $s$ you get
$$\sum_{i=1}^N c_if_{is}\frac{2d_i}{\sum_{j=1}^k \lambda_jf_{ij}}=b_s$$
If you had only one constraint, this would be
$$\sum_{i=1}^N c_i\frac{2d_i}{\lambda}=b$$
so that $$\lambda=\frac{2}{b}\sum_{i=1}^N c_id_i$$
and therefore $$x_i=\frac{bd_i}{f_{i}\sum_{m=1}^N c_md_m }.$$