3.1 Using KKT for Corner Solutions
In lecture 7 of Econ 50, you learned about “corner solutions,” in which the optimal solution lies at the boundary of the budget constraint; and in lecture 8, you learned about solutions at “kinks.” In both of these, the Econ 50 class was told that these were “situations in which calculus (i.e. Lagrange) didn’t work, so you needed to apply logic.”
This was not, in fact, true! We can actually use the KKT method we developed last week to multiple constraints, including both inequality constraints and equality constraints. And with this method, we can solve all the problems from lectures 7 and 8, plus a lot more.
In fact, the situation of corner solutions is very much like the example we saw last week in which you weren’t allowed to buy more than 4 units of good 1 ($x_1 \ge 4$). In fact, we always have two additional constraints for any budget constraint problem in which you can’t buy negative quantities: $x_1 \ge 0$ and $x_2 \ge 0$. We can therefore set up a Lagrangian using KKT with one equality two inequality constraints: \(\mathcal{L}(x_1,x_2,\lambda) = u(x_1,x_2) + \overbrace{\lambda(m - p_1x_1 - p_2x_2)}^{p_1x_1 + p_2x_2 = m} + \overbrace{\mu_1x_1}^{x_1 \ge 0} + \overbrace{\mu_2x_2}^{x_2 \ge 0}\) We can then proceed using the KKT as usual!
Worked example
Let’s see how this works with the example from lecture 7: maximizing the quasilinear utility function \(u(x_1,x_2) = 100 \ln x_1 + x_2\) subject to the budget constraint \(x_1 + 2x_2 = 100\)
We saw in the main class that the binding constraint was $x_2 \ge 0$:
because the unconstrained maximum (i.e. without the constraint $x_2 \ge 0$ would have occurred at $(200,-50)$:
Let’s see how to solve this with KKT. The Lagrangian is \(\mathcal{x_1,x_2,\lambda} = 100 \ln x_1 + x_2 + \lambda(100-x_1-2x_2) + \mu_1x_1 + \mu_2x_2\)
Notice that we will never choose to have strictly zero of good 1: indeed, since the utility derived from good 1 is $100\ln x_1$, having $x_1=0$ would imply infinitely negative utility! Thus we can save ourselves a step and conclude $x_1>0$ and so $\mu_1=0$.
From here, the first order conditions and complementary slackness conditions are: \(\frac{\partial \mathcal{L}}{\partial x_1} = \frac{100}{x_1} - \lambda + \mu_1 = 0\) \(\frac{\partial \mathcal{L}}{\partial x_2} = 1 - 2\lambda + \mu_2 = 0\) \(\frac{\partial \mathcal{L}}{\partial \lambda} = 100 - x_1 - 2x_2 = 0\) \(\mu_1 x_1 = 0, \qquad \mu_2 x_2 = 0\)
With $\mu_1 = 0$, the first condition gives $\lambda = \frac{100}{x_1}$. Suppose the constraint $x_2 \ge 0$ does not bind, so $\mu_2 = 0$. Then the second condition gives $\lambda = \tfrac{1}{2}$, so $x_1 = 200$ and the budget forces $x_2 = -50 < 0$, which is infeasible. We can therefore conclude the nonnegativity constraint binds and $x_2 = 0$.
With $x_2 = 0$, the budget gives $x_1 = 100$, hence $\lambda = \frac{100}{100} = 1$, and the second condition gives $\mu_2 = 2\lambda - 1 = 1 > 0$. The optimum is therefore \(x_1 = 100, \quad x_2 = 0, \quad \mu_1 = 0, \quad \mu_2 = 1.\) This is another good example of us “earning” some results from Econ 50. Not only does this formally verify that the solution without calculus is correct, but the values of those Lagrange multipliers are informative about marginal utilities in ways that help us quantify the “stress” on the solution, in particular how badly the consumer wants to break past each constraint. In future econ, engineering, or computer science courses, you’ll see how leveraging these multipliers can help solve more difficult optimization problems.