Maximization with lin-log objective

Question

Let $N = \{1,\ldots,n\}$. For a given $(r_1, \ldots, r_n) \in \mathbb R_{++}^n$. I need to solve \begin{align} \max_{(k_1, \ldots, k_n)\in \mathbb R_{++}^n} \prod_{i \in N}{\left[\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right]}. \end{align} I could not come up with a closed form solution of the maximizers by using first order conditions, because of the log. Is it possible to tell, that there does not exist a closed form solution and we thus have to maximize the product numerically.

Answer 1

$\ln$ is a strictly increasing monotonic function then with $x_k > 0$

$$ \max \prod_k x_x \equiv \max\sum_k \ln(x_k) $$

$$ \begin{align} \max_{(k_1, \ldots, k_n)\in \mathbb R_{++}^n} f(k)=\prod_{i \in N}{\left[\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right]}. \end{align} $$

so under the hypothesis that

$$ \sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right) > 0 $$

the problem presents a more amenable formulation as

$$ \max F(k) = \ln(f(k)) = \sum_{i\in N}\ln\left(\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right) $$

with

$$ \frac{\partial F}{\partial k_{\nu}} = \frac{r_{\nu}k_{\nu}-1}{\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right) }=0\Rightarrow r_{\nu} = k_{\nu} $$

and the stationary global point is attained at

$$ r_{\nu} = k_{\nu}, \forall\nu\in N $$