 gaiaecologicaq2

2022-07-07

Let $N=\left\{1,\dots ,n\right\}$. For a given $\left({r}_{1},\dots ,{r}_{n}\right)\in {\mathbb{R}}_{++}^{n}$. I need to solve
$\begin{array}{r}\underset{\left({k}_{1},\dots ,{k}_{n}\right)\in {\mathbb{R}}_{++}^{n}}{max}\prod _{i\in N}\left[\sum _{j\in N}\left({r}_{j}-{k}_{j}\right)+{r}_{i}\mathrm{ln}\left(\frac{{k}_{i}}{{r}_{i}}\right)\right].\end{array}$
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. Marisol Morton

Expert

$\mathrm{ln}$ is a strictly increasing monotonic function then with ${x}_{k}>0$
$max\prod _{k}{x}_{x}\equiv max\sum _{k}\mathrm{ln}\left({x}_{k}\right)$
$\begin{array}{r}\underset{\left({k}_{1},\dots ,{k}_{n}\right)\in {\mathbb{R}}_{++}^{n}}{max}f\left(k\right)=\prod _{i\in N}\left[\sum _{j\in N}\left({r}_{j}-{k}_{j}\right)+{r}_{i}\mathrm{ln}\left(\frac{{k}_{i}}{{r}_{i}}\right)\right].\end{array}$
so under the hypothesis that
$\sum _{j\in N}\left({r}_{j}-{k}_{j}\right)+{r}_{i}\mathrm{ln}\left(\frac{{k}_{i}}{{r}_{i}}\right)>0$
the problem presents a more amenable formulation as
$maxF\left(k\right)=\mathrm{ln}\left(f\left(k\right)\right)=\sum _{i\in N}\mathrm{ln}\left(\sum _{j\in N}\left({r}_{j}-{k}_{j}\right)+{r}_{i}\mathrm{ln}\left(\frac{{k}_{i}}{{r}_{i}}\right)\right)$
with
$\frac{\mathrm{\partial }F}{\mathrm{\partial }{k}_{\nu }}=\frac{{r}_{\nu }{k}_{\nu }-1}{\sum _{j\in N}\left({r}_{j}-{k}_{j}\right)+{r}_{i}\mathrm{ln}\left(\frac{{k}_{i}}{{r}_{i}}\right)}=0⇒{r}_{\nu }={k}_{\nu }$
and the stationary global point is attained at
${r}_{\nu }={k}_{\nu },\mathrm{\forall }\nu \in N$

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