B.4 Interpreting the Lagrange Multiplier

The Lagrange multiplier has an important intuitive meaning, beyond being a useful way to find a constrained optimum.

Let’s look at the Lagrangian for the fence problem again, but this time let’s assume that instead of 40 feet of fence, we have $F$ feet of fence. In this case the Lagrangian becomes \(\mathcal{L}(L,W,\lambda) = LW + \lambda(F - 2L - 2W)\) and the first-order conditions become \(\begin{aligned} \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial L} &= W - 2\lambda = 0 &\Rightarrow& W = 2\lambda \\ \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial W} &= L - 2\lambda = 0 &\Rightarrow& L = 2\lambda \\ \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial \lambda} &= F - 2L - 2W =0 &\Rightarrow& 2L + 2W = F \end{aligned}\) Solving this gives us our solutions \(\begin{aligned} L^\star &= {F \over 4}\\ W^\star &= {F \over 4}\\ \lambda &= {F \over 8} \end{aligned}\) What happens, though, if we take our optimal values of $L^\star$ and $W^\star$ and plug them back into the objective function $A(L,W) = LW$? In that case we can find the maximum possible area for any length of fence $F$: \(A^\star(F) = L^\star(F) \times W^\star(F) = {F \over 4} \times {F \over 4} = {F^2 \over 16}\) (Intuitively, all this is saying is that for any amount of perimeter $F$, the best we can do is to create a square with side lengths $F/4$, which has an area of $(F/4)^2 = F^2/16$.)

With this interpretation, note that $\lambda$ represents the derivative of this $A^\star(F)$ function: \({dA^\star(F) \over dF} = {d \over dF}\left({F^2 \over 16}\right) = {F \over 8}\) Visually, we can see this as the slope of $A^\star(F)$:

In other words, $\lambda$ tells us the amount by which the objective function rises due to a one-unit relaxation of the constraint. (We can also see that if we take the derivative of the Lagrangian with respect to $F$, we get $\lambda$.)

In economics, this value of $\lambda$ is often called a “shadow price.” For example, in consumer theory, we’ll use the Lagrange multiplier method to maximize utility given a constraint defined by the amount of money, $m$, you have to spend; the value of $\lambda$ in that problem will yield the additional utiltiy you’d get from getting another dollar to spend. Likewise, in producer theory, we’ll use the Lagrange method to solve for the cost-minimizing combination of labor and capital required to produce some amount of output, $q$; the value of $\lambda$ in that problem will be the marginal cost of producing an additional unit of output.

In practice, we can often solve constrained optimization problems without directly invoking a Lagrange multiplier. However, it’s important to understand the critical role this multiplier plays behind the scenes.