The Lagrange Method of Finding an Optimal Bundle

We’ve established that under certain circumstances, the solution to the consumer’s problem $\begin{aligned} \max_{x_1,x_2}\ & u(x_1,x_2)\\ \text{s. t. }\ & p_1x_1 + p_2x_2 \le m \end{aligned}$ is characterized by two equations: $\begin{aligned}\text{Tangency condition:\ \ \ } & {MU_1 \over MU_2} = {p_1 \over p_2}\\\text{ Budget constraint:\ \ \ } & p_1x_1 + p_2x_2 = m\end{aligned}$ In fact, we can see that this is the same result as if we solved the problem using the method of Lagrange multipliers. Furthermore, we can see that the Lagrange multiplier itself has an interesting economic interpretation.

The 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.” Therefore we can think about the consumer maximizing this Lagrangian as balancing the money spent on good 1, the money spent on good 2, money spent on other things.

Where does the Lagrange multiplier come into play here? Well, since $u(x_1,x_2)$ is measured in utils, and $m - p_1x_1 - p_2x_2$ is measured in dollars we can’t just add them together; we need to measure everything in the same units. The Lagrange multiplier $\lambda$, which is measured in utils per dollar, serves as a kind of “exchange rate” between utility and money.

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 we developed here, we can see that $\lambda$ must be equal to the “bang for the buck” from the last unit of either good you spend money on. 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 tangency condition and the budget constraint, just as we derived before!

More than two goods

Note that the Lagrange solution works with any number of goods, though, not just two. That is, if we have $n$ goods, the problem we’d be solving would be $\begin{aligned} \max_{x_1,x_2}\ & u(x_1,x_2,x_3,\cdots,x_n)\\ \text{s. t. }\ & p_1x_1 + p_2x_2 + p_3x_3 + \cdots + p_nx_n \le m \end{aligned}$the Lagrangian would be$\begin{aligned}\mathcal{L}(x_1,x_2,x_3, \cdots, x_n,\lambda) &= u(x_1,x_2,x_3, \cdots, x_n)\\ & + \lambda(m - p_1x_1 - p_2x_2 - p_3x_3 - \cdots - p_nx_n)\end{aligned}$and 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 when the consumer is optimizing; otherwise, they could reallocate their money and get more utility.

Just one good

This section is purely optional, but I think it’s neat so I’m including it.

In fact, the Lagrange method can be used with only one good. Here the solution is trivial, but the interpretation of the Lagrange multiplier is perhaps even clearer than it would otherwise be.

Suppose you’re trying to maximize $\begin{aligned} \max_{x}\ & u(x)\\ \text{s. t. }\ & px \le m \end{aligned}$ If $u’(x) > 0$ (that is, if more of good 1 is always better), then the answer is clearly just “spend $m$ on good 1.”

But let’s see how the Lagrangian works here. We have $\mathcal{L}(x,\lambda) = u(x) + \lambda(m - px)$ Look at what happens if we plot this Lagrangian. If $\lambda = 0$, we just have the utility function. As we increase $\lambda$, the value at the point along the constraint $x=m/p$ (showed by the green vertical line) remains the same, but the slope gets flatter and flatter. When $\lambda$ is exactly equal to the utility gain from spending your last dollar on good 1, the Lagrangian is flat at the optimal point:

One way of thinking about the Lagrangian is that it punishes the consumer for spending money: as $x$ increases, $\mathcal{L}(x,\lambda)$ increases by $u^\prime(x)$ but decreases by $\lambda p$: that is, each additional unit gives you some amount of marginal utility, but it costs $p$ dollars, and each dollar you don’t have any more costs you $\lambda$ utils. Furthermore, if you were to get just one more dollar, your utility would increase by (approximately) $\lambda$.