2.3 Inequality Constraints with Two Choice Variables

Let’s work through an example of an inequality constraint in an economic context with two choice variables, one equality constraint (the budget line), and one inequality constraint. Specifically, let’s think about a problem in which for some reason you can’t consume more than a certain amount of good 1.

We’ll build off the utility maximization problem from lecture 6: maximizing the utility function $u(x_1,x_2) = x_1^{3 \over 4}x_2^{1 \over 4}$ subject to the constraint $4x_1 + 2x_2 = 48$. Let’s make our lives a little easier mathematically and transform the utility function (by taking the natural log and multiplying by 4) to be \(u(x_1,x_2) = 3 \ln x_1 + \ln x_2\) Therefore the Lagrangian for this problem is \(\mathcal{L}(x_1,x_2,\lambda) = 3 \ln x_1 + \ln x_2 + \lambda(48 - 4x_1 - 2x_2)\) which gives us a solution at $(9,6)$:

Let’s now impose an inequality constraint on this problem: let’s say that this store limits purchases of good 1 to 4 units per customer: that is, $x_1 \le 4$. Let’s solve for the constrained optimal bundle using the KKT method, including the Lagrange multipliers $\lambda$ and $\mu$.

Step 1: Set up the problem \(\begin{aligned} \text{objective:} \quad & u(x_1,x_2) = 3 \ln x_1 + \ln x_2\\ \text{equality constraint:} \quad & g(x_1,x_2) = 48 - 4x_1 - 2x_2 = 0\\ \text{inequality constraint:} \quad & h(x_1,x_2) = 4 - x_1 \geq 0 \end{aligned}\)

Step 2: Write the Lagrangian \(\begin{aligned} \mathcal{L}(x_1,x_2,\lambda,\mu) &= u(x_1,x_2) + \lambda g(x_1,x_2) + \mu h(x_1,x_2)\\ &= 3 \ln x_1 + \ln x_2 + \lambda(48 - 4x_1 - 2x_2) + \mu(4 - x_1)\end{aligned}\)

Step 3: Write complementary slackness \(\mu(4 - x_1) = 0\) This is equivalent to saying either $\mu=0$ (that is, the constraint is slack and does not bind), or $(4-x_1)=0$ (which implies $x_1=4$ and so the constraint binds). We do not yet know which will be the case.

Step 4: First-order conditions \(\begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} &= \frac{3}{x_1} - 4\lambda - \mu = 0\\ \\ \frac{\partial \mathcal{L}}{\partial x_2} &= \frac{1}{x_2} - 2\lambda = 0\\ \\ \frac{\partial \mathcal{L}}{\partial \lambda} &= 48 - 4x_1 - 2x_2 = 0\end{aligned}\)

Step 5: Solve each case. We need to check whether the inequality constraint will affect our choice or not. The most straightforward approach I like to use is to assume the constraint does not bind, look at the implied solution, and see whether we violate the inequality constraint.

Case 1: $\mu = 0$ (constraint slack): We can start by plugging $\mu=0$ into the first order conditions and solving for $\lambda$. This gives us just the same tangency condition as we had in the readings for lecture 6: \(\frac{3}{x_1} = 4\lambda, \quad \frac{1}{x_2} = 2\lambda \implies x_2 = {2 \over 3}x_1\) As in that reading, we then plug the result into the budget constraint to find the optimum at $(9,6)$. However, this violates the constraint $x_1 \leq 4$, so we conclude $\mu$ must not be zero – that is, the inequality constraint $x_1 \le 4$ must bind.

Case 2: $\mu > 0$ (constraint binds): If the constraint binds, we know that $x_1 = 4$. This means our first-order conditions are now: \(\begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} &= \frac{3}{4} - 4\lambda - \mu = 0\\ \\ \frac{\partial \mathcal{L}}{\partial x_2} &= \frac{1}{x_2} - 2\lambda = 0\\ \\ \frac{\partial \mathcal{L}}{\partial \lambda} &= 48 - 4 \times 4 - 2x_2 = 0\end{aligned}\) The last equation tells us $x_2 = 16$; plugging this into the second gives us $\lambda = 1/32$; plugging that into the first gives us $\mu = 5/8$. Since $\mu > 0$ is consistent with a binding constraint, this is the solution.

How do we interpret the two multipliers? Think about what would happen if each one were relaxed by a bit. The multiplier $\lambda$ measures the effect of the budget constraint; it says how much additional utility you’d get if you had another dollar. In this case, if you had another dollar, you would have to spend it on good 2, since you’re already buying the maximum possible good 1. Since $p_2 = 2$, this means you could buy 1/2 unit of good 2, which would increase your utility by \(\Delta u = {1 \over p_2} \times MU_2(4,16) = {1 \over 2} \times {1 \over 16} = {1 \over 32}\) This is, indeed, what we found for $\lambda$!

On the other hand, if the restriction $x_1 \le 4$ was relaxed by one, you would buy one more unit of good 1, and two fewer units of good 2: therefore your change in utility would be \(\Delta u = 1 \times MU_1(4,16) - 2 \times MU_2(4,16) = {3 \over 4} - {2 \over 16} = {5 \over 8}\) which is the value we found for $\mu$.

The diagram below shows how $x_1, x_2, \lambda,$ and $\mu$ change as the restriction $x_1 \le K$ changes. Before you play with it: what do you think happens to $\mu$ if $K > 9$?