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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.

  1. 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?

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

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

Thanks!

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  • $\begingroup$ Why are you adding $\mu$'s to the lagrangian, and why are you differentiating lagrangian with respect to them? Are they some additional constraints? $\endgroup$ – 1muflon1 Oct 31 '20 at 0:47
  • $\begingroup$ Yea I thought since $x_1 \geq 0$ and $x_2 \geq 0$ were natural constraints. Obviously, they can't actually equal 0 here, but if I change the problem to like just the term $x_1$ rather than $\ln(x_1)$ then it can equal 0. $\endgroup$ – Alex Oct 31 '20 at 0:50
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    $\begingroup$ These non-negativity constraints are superfluous given that there is already a positive budget constraint. With inequality constraints you will also have to check the Kuhn-Tucker conditions and it would have to be solved slightly differently. The reason why often lagrangian is for a budget applied with equality even if there is an inequality condition there is that rational person will always consume at the budget line but you cant use the same argument to replace the non-negativity constraints with equality to zero. $\endgroup$ – 1muflon1 Oct 31 '20 at 1:03
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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 $x_i > 0$, by the complimentary slackness condition, $\mu_i = 0$.

See Karush-Kuhn-Tucker conditions. It also contains regularity conditions you ask for.

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  • $\begingroup$ Thanks! So I all have to say is clearly from conditions 1 and 2, $x_i > 0$ because otherwise if $x_1 = 0$, then that implies $\lambda = \infty$ and $x_2 = 0$, so we have a contradiction from the third condition (and similarly for $x_2$)? Or can I just say clearly from the original utility function, we cannot have $x_i = 0$ since you cannot take the natural log of 0. $\endgroup$ – Alex Oct 31 '20 at 1:14
  • $\begingroup$ I think the first argument works. $\endgroup$ – Walrasian Auctioneer Oct 31 '20 at 1:56

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