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Chapter 7 / Utility Maximization Subject to a Budget Constraint

7.7 Appendix: Interpreting the Lagrange Conditions for a Utility Maximization Problem


The consumer’s constrained utility maximization problem is \(\begin{aligned} \max_{x_1,x_2}\ & u(x_1,x_2)\\ \text{s. t. }\ & p_1x_1 + p_2x_2 \le m \end{aligned}\) The corresponding Lagrangian for this problem is: \(\mathcal{L}(x_1,x_2,\lambda) = u(x_1,x_2) + \lambda(m - p_1x_1 - p_2x_2)\) Note that since $p_1x_1$ is the amount of money spent on good 1, and $p_2x_2$ is the amount of money spent on good 2, we can interpret $m - p_1x_1 - p_2x_2$ as “money left over to spend on other things.” Since $u(x_1,x_2)$ is measured in utils, and $m - p_1x_1 - p_2x_2$ is measured in dollars, it must be the case that the Lagrange multiplier $\lambda$ is measured in utils per dollar.

To find the optimal bundle, we take the first-order conditions of this Lagrangian with respect to the choice variables $x_1$ and $x_2$ and the Lagrange multiplier $\lambda$: \(\begin{aligned} {\partial \mathcal{L} \over \partial x_1} &= MU_1 - \lambda p_1 = 0\\ {\partial \mathcal{L} \over \partial x_2} &= MU_2 - \lambda p_2 = 0\\ {\partial \mathcal{L} \over \partial \lambda} &= m - p_1x_1 - p_2x_2 = 0 \end{aligned}\) Solving the first two FOC’s for $\lambda$ gives us \(\lambda = {MU_1 \over p_1} = {MU_2 \over p_2}\) Using the interpretation from above, this is saying that the “bang for the buck” from the last unit of good 1 must be the same as the “bang for the buck” from the last unit of good 2; and that both of these are equal to the Lagrange multiplier $\lambda$. In other words, $\lambda$ is not just measured in units of utils per dollar: when the consumer is optimizing, it represents the additional utility the consumer could get by spending an additional dollar on either good. We can also see this by noting that the derivative of the Lagrangian with respect to $m$ is $\lambda$: \({d \over dm}\left[u(x_1,x_2) + \lambda(m - p_1x_1 - p_2x_2)\right] = \lambda\) That is, if we gave the consumer one more dollar, their utility would increase by approximately $\lambda$.

Cross-multiplying, we can see that the above condition is simply \({MU_1 \over MU_2} = {p_1 \over p_2}\) which is the “tangency condition” $MRS = p_1/p_2$. The third FOC, of course, just gives us our budget constraint back. Therefore the end result of the Lagrange method may be characterized by the two conditions that we saw in the last section!

Note that the Lagrange solution works with any number of variables, though, not just two. That is, if we have $n$ goods, the Lagrangian would be \(\mathcal{L}(x_1,x_2,x_3, \cdots, x_n,\lambda) = u(x_1,x_2,x_3, \cdots, x_n) + \lambda(m - p_1x_1 - p_2x_2 - p_3x_3 - \cdots - p_nx_n)\) the first order conditions would be \(\begin{aligned} {\partial \mathcal{L} \over \partial x_1} &= MU_1 - \lambda p_1 = 0\\ {\partial \mathcal{L} \over \partial x_2} &= MU_2 - \lambda p_2 = 0\\ {\partial \mathcal{L} \over \partial x_3} &= MU_3 - \lambda p_3 = 0\\ \vdots \\ {\partial \mathcal{L} \over \partial x_n} &= MU_n - \lambda p_n = 0\\ {\partial \mathcal{L} \over \partial \lambda} &= m - p_1x_1 - p_2x_2 - p_3x_3 - \cdots - p_nx_n = 0 \end{aligned}\) and the first $n$ conditions would resolve to \(\lambda = {MU_1 \over p_1} = {MU_2 \over p_2} = {MU_3 \over p_3} = \cdots = {MU_n \over p_n}\) This system of equations can then be used to solve for the optimal bundle, no matter how many goods are involved. However, the intuition remains the same: the “bang for the buck” from every good, represented by $\lambda$, must be equal.

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