I want to solve this production function using Lagrange multiplier
$$\begin{equation} \begin{aligned} \max_{} \quad & U(x)=(\sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}})^{\frac{1}{\rho}} \\ \textrm{s.t.} \quad & \sum_{i=1}^{n}{p_{i}x_{i}=I}\\ \end{aligned} \end{equation}$$
What I have been doing
Apply lagrangian
$$L(x_i, \lambda)=(\sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}})^{\frac{1}{\rho}} - \lambda(\sum_{i=1}^{n}{p_{i}x_{i} - I})$$
Finding the FOCs
$$\frac{\partial{L}}{\partial{\lambda}} = I - \sum_{i=1}^{n}{p_{i}x_{i}}$$ $$\frac{\partial{L}}{\partial{x_i}} = \frac{\theta_k x_k^\rho(\sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}})^{\frac{1}{\rho}})}{x_k(\sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}})} - \lambda p_k$$ Process here
Why can't it be solved this way?
I've been reading and found that in order to solve this problem you have to find a growing monotone function to simplify the utility function.
$$v(x)=U(x)^\rho $$ $$\max_{} \quad v(x)=\sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}}$$
There would be how to work with the long line?
In this case, maximizing $U(x)$ and $U(x)^p$ would be equivalent. The second case is much easier though: if you write the problem \begin{equation} \begin{aligned} \max_{} \quad & U(x)^p = \sum_{i=1}^{n}{\theta_{i}x_{i}^{\rho}} \\ \textrm{s.t.} \quad & \sum_{i=1}^{n}{p_{i}x_{i}=I}\\ \end{aligned} \end{equation} then the optimality conditions are:
\begin{equation} \begin{aligned} \theta_i \rho x_i^{\rho-1} - \lambda p_i = 0 \quad \forall i \\ \sum_{i=1}^n p_i x_i = I \end{aligned} \end{equation} Injecting the first one into the second one, we get \begin{equation} \sum_{i=1}^n p_i \left(\frac{\lambda p_i}{\theta_i} \right)^{\frac{1}{\rho-1}} = I \end{equation} then solve for $\lambda$ and inject back in the first equation.