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?

  • $\begingroup$ $\;x_1,\,x_2>0\;$ 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

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 $$ \lambda_1 (-1,0) + \lambda_2 (-1,-1) = - \left(\frac{1}{w_1},\frac{\beta}{w_2}\right) $$


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