9.6 Appendix A: Cost Minimization with Lagrange

Utility maximization and cost minimization are both constrained optimization problems of the form \(\begin{aligned} \max_{x_1,x_2}\ \ \ & f(x_1,x_2) \\ \text{s.t.}\ \ \ & g(x_1,x_2) = 0 \end{aligned}\) In this kind of constrained optimization problem we call the function $f(x_1,x_2)$ the objective function and the equation $g(x_1,x_2) = 0$ the constraint.

Under certain conditions, we may find the solution to a constrained optimization problem by setting up the Lagrangian \(\mathcal{L}(x_1,x_2,\lambda) = f(x_1,x_2) + \lambda g(x_1,x_2)\) The first-order conditions set the partial derivatives of this equal to zero, giving us \(\begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} = \frac{\partial f}{\partial x_1} + \lambda \frac{\partial g}{\partial x_1} &= 0 \Rightarrow \lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1}\\ \frac{\partial \mathcal{L}}{\partial x_2} = \frac{\partial f}{\partial x_2} + \lambda \frac{\partial g}{\partial x_2} &= 0 \Rightarrow \lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2}\\ \frac{\partial \mathcal{L}}{\partial \lambda} = g(x_1,x_2) &= 0 \end{aligned}\) Setting equations (1) and (2) equal to one another gives us the “tangency condition” \(\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{\partial f/\partial x_2}{\partial g/\partial x_2}\) or \(\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2}\) So the solution here may be described as “the point along the constraint where the tangency condition is met.”

For what we’ve seen thus far, the objective function has been the “utility function” whose output is measured in utils, and the constraint has been the budget constraint: that is, \(\begin{aligned} \textbf{Objective function: }f(x_1,x_2) &= u(x_1,x_2)\\ \textbf{Constraint (when set equal to zero): }g(x_1,x_2) &= m - p_1x_2 - p_2x_2 \end{aligned}\) Mathematically, we can see that \(\frac{\partial f}{\partial x_1} = MU_1\) \(\frac{\partial f}{\partial x_2} = MU_2\) \(\frac{\partial g}{\partial x_1} = -p_1\) \(\frac{\partial g}{\partial x_2} = -p_2\) so our values for $\lambda$ in equations (1) and (2) have been \(\lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{MU_1}{p_1}\) \(\lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2} = \frac{MU_2}{p_2}\) We called this the “bang for the buck” condition: that is, $\lambda$ represented the utility gain from the last dollar spent on either good, or more generally the utility gained if you were to get one more dollar of income.

Setting these two values of $\lambda$ equal to one another, our tangency condition has been \(\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2} \Rightarrow \frac{MU_1}{MU_2} =\frac{p_1}{p_2}\) This would give us one relationship between good 1 and good 2; the constraint would give us another; and the solution could be found at the intersection of those two conditions. In other words, the optimal bundle is the one along the budget constraint where the tangency condition holds.

For the cost minimization problem, we swap the objective function and the constraint, which is now given by an exogenous level of income $U$: \(\begin{aligned} \min_{x_1,x_2}\ \ \ & p_1x_1 + p_2x_2 \\ \text{s.t.}\ \ \ & u(x_1,x_2) = U \end{aligned}\) Now, using the same $f()$ and $g()$ framing as above, we have \(\begin{aligned} \textbf{Objective function: }f(x_1,x_2) &= p_1x_2 + p_2x_2\\ \textbf{Constraint (when set equal to zero): }g(x_1,x_2) &= U - u(x_1,x_2) \end{aligned}\)

Mathematically, following the same procedure as above, we have \(\frac{\partial f}{\partial x_1} = p_1\) \(\frac{\partial f}{\partial x_2} = p_2\) \(\frac{\partial g}{\partial x_1} = -MU_1\) \(\frac{\partial g}{\partial x_2} = -MU_2\) so our values for $\lambda$ in equations (1) and (2) are now \(\lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{p_1}{MU_1}\) \(\lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2} = \frac{p_2}{MU_2}\) Rather than being “bang for the buck” (measured in utils per dollar), these could be thought of as “buck for the bang” (measured in dollars per util) – that is, $\lambda$ now measures the cost, in dollars, of increasing one’s utility by 1.

Of course, setting the two values of $\lambda$ equal to one another give us the exact same tangency condition as before: \(\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2} \Rightarrow \frac{p_1}{p_2} = \frac{MU_1}{MU_2}\) As above, this would give us one relationship between good 1 and good 2; the constraint would give us another; and the solution could be found at the intersection of those two conditions. Because the constraint is not a budget constraint, but rather a utility constraint – that is, we’re constraining ourselves to be along the indifference curve representing bundles , the optimal bundle is the one along the utility constraint where the tangency condition holds.

How do we think about the duality of this relationship? Well, the tangency condition – or more generally, the IOC – shows various optimal points, in the sense that along that line, there is no overlap between the set of bundles preferred to a point and the set of bundles cheaper than the point. Hence it doesn’t really matter if we think of each point along the IOC as the most effective use of a fixed budget, or the cheapest way to afford a fixed amount of utility. But if we look at a series of indifference curves and budget/iso-cost lines, we can’t tell which kind of problem we’re solving:

We could be maximizing utility subject to four budget constraints, or we could be minimizing cost subject to four utility constraints. Either way, the solution lies at the intersection of the tangency condition and the constraint.