Solving Utility Maximization with Lagrangian I know how to solve the 2 variable constrained optimization problem using MRS = MRT, but I also want to make sure I understand how to do it with the Lagrangian method.
So if I have the following problem
$U(x) = \alpha\ln(x_1) + (1-\alpha)\ln(x_2)$
with $p_1x_1 + p_2x_2 = w$
I got the answer using the MRS = MRT method as $x_1 = \frac{w\alpha}{p_1}$ and $x_2 = \frac{w(1-\alpha)}{p_2}$.  I am a bit confused on how to set up the Lagrangian.  Here's what I did
So $L = \alpha\ln(x_1) + (1-\alpha)\ln(x_2) + \lambda(w - p_1x_1 - p_2x_2) + \mu_1x_1 + \mu_2x_2$
$\frac{dL}{dx_1} = \frac{\alpha}{x_1} + p_1\lambda + \mu_1 = 0$
$\frac{dL}{dx_1} = \frac{1-\alpha}{x_2} + p_2\lambda + \mu_2 = 0$
$\frac{dL}{d\lambda} = w - p_1x_1 - p_2x_2 = 0$
$\frac{dL}{d\mu_1} = x_1 = 0$
$\frac{dL}{d\mu_2} = x_2 = 0$
Here is my issue here.  If I assume $x_1$ and $x_2$ cannot be 0 and I somehow assume $\mu_1$ and $\mu_2$ are 0, then I can solve it pretty easily.  I then just equate the $\lambda$ in the first two equations and then plug into the budget constraint like in the MRS = MRT case.

However, what gives me the right to make $\mu_1$ and $\mu_2$ equal to 0?  Is this the correct approach?  When are they not 0?

I heard in order to use the Lagrangian method, some &quot;conditions&quot; need to be satisfied.  What conditions need to be satisfied?  How do I verify this?  Is this related to differentiability?

Is there some restrictions on what $\lambda$ and $\mu$ can be?


Thanks!
You have added inequality constraints $x_1 \geq 0$ and $x_2 \geq 0$.
Hence, you now have Kuhn-Tucker conditions for complimentary slackness:
$$
\mu_1 x_1 = 0 \text{ and } \mu_2x_2 = 0
$$
So whenever $x_i &gt; 0$, by the complimentary slackness condition, $\mu_i = 0$.
See Karush-Kuhn-Tucker conditions. It also contains regularity conditions you ask for.