# Interpretation of this Lagrange Multiplier

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?

• $\;x_1,\,x_2>0\;$ 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... – DonAntonio Jul 13 '19 at 12:57
• @DonAntonio I have already solved for all of the unknowns, but my point of confusion is the interpretation of $\lambda_4$. – David Jul 13 '19 at 13:18
• I have edited the problem to not include the strict positiveness constraints on $x_1$ and $x_2$, which were redundant. – David Jul 13 '19 at 13:32
• That Langranian is just valid for when the epxressions are equalities ...so solve that, and then solve of all the values inside that domain. – DonAntonio Jul 13 '19 at 14:14
• @DonAntonio I don’t think you’re understanding me. I’ve already solved the unknowns. Look at the question I’m asking please. – 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 $$\lambda_1 (-1,0) + \lambda_2 (-1,-1) = - \left(\frac{1}{w_1},\frac{\beta}{w_2}\right)$$