0
$\begingroup$

I have the following utility maximization problem with inequality constraints:

Objective function given by $U(x_1,x_2)=\ln(x_1)+\beta \ln(x_2)$ where $0<\beta<1$, and the constraints are given by $0\leq w_1 - x_1$ and $0\leq w_1+w_2-x_1-x_2$, where $w_1$ and $w_2$ are strictly positive.

Because the objective function and constraint functions are concave and differentiable, we can use Kuhn-Tucker and find the $(x_1, x_2)$ pair that solves the first-order conditions of the Lagrangian.

With the Lagrangian expressed as: $$L=U(x_1, x_2)+\lambda_1(w_1-x_1)+\lambda_2(w_1+w_2-x_1-x_2),$$

I have determined that $\lambda_1=\frac{1}{w_1}-\frac{\beta}{w_2}$ and $\lambda_2=\frac{\beta}{w_2}$. My question is, how should I interpret the value of $\lambda_2$?

My current belief is that it represents the increase in the maximum utility when $w_1+w_2$ increases by $1$, but I'm confused by the fact that if $w_1$ were to increase, it would also involve $\lambda_1$, unless only $w_2$ increased. Can someone help to clarify?

$\endgroup$
  • $\begingroup$ $\;x_1,\,x_2>0\;$ are a must, and I wouldn't call them "constraints" , and $\;0<\beta<1\;$ is just a parameter, which is also not a constraint imo. About the two inequalities: solve the problem as one-variable max-min usual problem, which requires differentiability and thus $\;w_1-x_1>0\,,\,\,w_1+w_2>x_1+x_2\;$ . and after that assumme equality ($\,x_1=w_1\,,\,x_2=w_2\;$) and just substitute and compare all the obtained values... $\endgroup$ – DonAntonio Jul 13 '19 at 12:57
  • $\begingroup$ @DonAntonio I have already solved for all of the unknowns, but my point of confusion is the interpretation of $\lambda_4$. $\endgroup$ – David Jul 13 '19 at 13:18
  • $\begingroup$ I have edited the problem to not include the strict positiveness constraints on $x_1$ and $x_2$, which were redundant. $\endgroup$ – David Jul 13 '19 at 13:32
  • $\begingroup$ That Langranian is just valid for when the epxressions are equalities ...so solve that, and then solve of all the values inside that domain. $\endgroup$ – DonAntonio Jul 13 '19 at 14:14
  • $\begingroup$ @DonAntonio I don’t think you’re understanding me. I’ve already solved the unknowns. Look at the question I’m asking please. $\endgroup$ – David Jul 13 '19 at 14:16
0
$\begingroup$

With $\beta > 0$ the maximum will be located at the boundary. So the potential maximum point is $w_1,w_2$. At this point we have

$$ \nabla (w_1-x_1) = (-1,0)\\ \nabla (w_1+w_2-x_1-x_2) = (-1,-1)\\ \nabla (\ln x_1 + \beta\ln x_2) = \left(\frac{1}{w_1},\frac{\beta}{w_2}\right) $$

so the condition for a maximum is the existence of $\lambda_1\ge 0 ,\lambda_2 \ge 0$ such that $$ \lambda_1 (-1,0) + \lambda_2 (-1,-1) = - \left(\frac{1}{w_1},\frac{\beta}{w_2}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.