Mathematical methods for economic theory

Martin J. Osborne
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6.1.2 Optimization with an equality constraint: interpretation of Lagrange multipliers

Consider the problem
maxx,y f(x, y) subject to g(x, y) = c,
where f and g are defined on the domain S.

Suppose we solve the problem for some value of c. Denote the solution (x*(c), y*(c)). Assume that g'1(x*(c), y*(c)) ≠ 0 or g'2(x*(c), y*(c)) ≠ 0 and that (x*(c), y*(c)) is an interior point of S. Then by a previous result there is a number λ*(c) such that (x*(c), y*(c), λ*(c)) satisfies the first-order conditions given in the result.

Assume that the functions x*, y*, and λ* are differentiable and let f*(c) = f(x*(c), y*(c)). Differentiate f*:

f*'(c)  =  f'x(x*(c), y*(c))x*'(c) + f'y(x*(c), y*(c))y*'(c)
 =  λ*(c)[g'x(x*(c), y*(c))x*'(c) + g'y(x*(c), y*(c))y*'(c)]
(using the first-order conditions). But g(x*(c), y*(c)) = c for all c, so the derivatives of each side of this equality are the same. That is,
g'x(x*(c), y*(c))x*'(c) + g'y(x*(c), y*(c))y*'(c) = 1 for all c.
Hence
f*'(c) = λ*(c).

That is,

if the solution of the problem is an interior point of the domain of the objective function that satisfies the first-order conditions and is a differentiable function of c, then the value of the Lagrange multiplier at the solution is equal to the rate of change in the maximal value of the objective function with respect to c.
For example, in a utility maximization problem the value of the Lagrange multiplier measures the marginal utility of income: the rate of increase in maximized utility as income increases.

Example 6.1.2.1
Consider the problem

maxxx2 subject to x = c.
The solution of this problem is obvious: x = c (the only point that satisfies the constraint!). The maximized value of the function is thus c2, so that the derivative of this maximized value with respect to c is 2c.

Let's check that the value of the Lagrange multiplier at the solution of the problem is equal to 2c. The Lagrangean is

L(x) = x2 − λ(x − c),
so the first-order condition is

2x − λ = 0.
The constraint is x = c, so the pair (x, λ) that satisfies the first-order condition and the constraint is (c, 2c). Thus we see that indeed λ is equal to the derivative of the maximized value of the function with respect to c.

Example 6.1.2.2
A firm uses two inputs to produce one output. Its production function is

f(x, y) = xayb.

The price of output is p, and the prices of the inputs are wx and wy. The firm is constrained by a law that says it must use exactly the same number of units of both inputs.

Thus the firm's problem is

maxx,y [pxayb − wxx − wyy] subject to y − x = 0.

(The firm is also constrained by the conditions x ≥ 0 and y ≥ 0, but I am ignoring these constraints at the moment.)

The Lagrangean is

L(x,y) = pxayb − wxx − wyy − λ(y − x)
so the first-order conditions are
apxa−1yb − wx + λ  =  0
bpxayb−1 − wy − λ  =  0
and the constraint is y = x. These equations have a single solution, with
x = y = ((wx + wy)/(p(a + b)))1/(a+b−1)
and
λ = (bwx − awy)/(a + b).
There is no value of (x, y) for which g'1(x, y) = g'2(x, y) = 0, so if the problem has a solution it is the solution of the first-order conditions.

Since λ measures the rate of increase of the maximal value of the objective function as the constraint is relaxed, it follows that if λ > 0 then the firm would like the constraint to be relaxed: its profit would be higher if the constraint were y − x = ε, for some ε > 0.

Suppose that bwx > awy, so that λ > 0, and the firm would like to use more of input y than of input x. A government inspector indicates that for a bribe, she is willing to overlook a small violation of the constraint: she is willing to allow the firm to use a small amount more of input y than it does of input x. Suppose the constraint is relaxed to y − x = ε. The maximum bribe the firm is willing to offer is the increase in its maximized profit, which is approximately ελ = ε(bwx − awy)/(a + b). Hence this is the maximum bribe the firm is willing to pay. (If wx = wy = 1, a = 1/4, and b = 1/2, for example, the maximum bribe is ε/3.)