This is a microeconomics question, however, as it is the calculations that cause problems I hope you can help me anyways.
I have to minimize the following using Lagrange:
$$ \min p_1x_1+p_2x_2 $$ subject to $(x_1^p + x_2^p)^{1/p} \geq U$.This means that I have to find $x_1$ and $x_2$ where $x_1$ and $x_2$ may not depend on each other. Thus let $$ L = p_1x_1+p_2x_2 - \lambda \left( \left(x_1^p+x_2^p \right)^{1/p} - U \right) $$ We get that \begin{align} \frac{\partial L}{\partial x_1} := p_1 - \lambda \left( \frac{1}{p} \left(x_1^p + x_2^p \right)^{(1/p)-1} px_1^{p-1} \right) = 0 \end{align}
\begin{align} \frac{\partial L}{\partial x_2} := p_1 - \lambda \left( \frac{1}{p} \left(x_1^p + x_2^p \right)^{(1/p)-1} px_2^{p-1} \right) = 0 \end{align}
\begin{align} \frac{\partial L}{\partial \lambda} := (x_1^p + x_2^p)^{1/p} = U \end{align}
Dividing the first expression with expression 2 one gets $$ \frac{p_1}{p_2} = \frac{x_1^{p-1}}{x_2^{p-1}} $$ Thus $$ x_1 = \left( \frac{p_1x_2^{p-1}}{p_2} \right)^{1/(p-1)} $$ and then I have to substitute this expression in the third expression but I simply can't see a way to solve for $x_2$ then. Can you help me?
I did not go through your computations. But from $$ \frac{p_1}{p_2} = \frac{x_1^{p-1}}{x_2^{p-1}} $$ you can deduce that $$x_1=\lambda\>p_1^{1/(p-1)},\qquad x_2=\lambda\>p_2^{1/(p-1)}$$ for a certain $\lambda>0$. The condition $\left(x_1^p+x_2^p\right)^{1/p}=U$ then determines $\lambda$: $$\lambda\left(p_1^{p/(p-1)}+p_2^{p/(p-1)}\right)^{1/p}=U\ .$$