Production function problem (Lagrange multiplier)

Question

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?

cvanaret · Answer

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.

Production function problem (Lagrange multiplier)

What I have been doing

Apply lagrangian

Finding the FOCs

Why can’t it be solved this way?

One Answer

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