6.1.2 Optimization with an equality constraint: interpretation of Lagrange multipliers
Let f*(c) = f(x*(c), y*(c)). Differentiate f*(c) with respect to c:
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)] 
That is,
the value of the Lagrange multiplier at the solution of the problem is equal to the rate of change in the maximal value of the objective function as the constraint is relaxed.
 Example

Consider the problem
max_{x}x^{2} 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 c^{2}, 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) = x^{2} − λ(x − c),so the firstorder condition is2x − λ = 0.The constraint is x = c, so the pair (x, λ) that satisfies the firstorder 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

A firm uses two inputs to produce one output. Its production function is
f(x, y) = x^{a}y^{b},
The price of output is p, and the prices of the inputs are w_{x} and w_{y}. 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
max_{x,y} [px^{a}y^{b} − w_{x}x − w_{y}y] 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) = px^{a}y^{b} − w_{x}x − w_{y}y − λ(y − x)so the firstorder conditions areapx^{a−1}y^{b} − w_{x} + λ = 0 bpx^{a}y^{b−1} − w_{y} − λ = 0 x = y = ((w_{x} + w_{y})/(p(a + b)))^{1/(a+b−1)}andλ = (bw_{x} − aw_{y})/(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 firstorder condition.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 bw_{x} > aw_{y}, 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(bw_{x} − aw_{y})/(a + b). Hence this is the maximum bribe the firm is willing to pay. (If w_{x} = w_{y} = 1, a = 1/4, and b = 1/2, for example, the maximum bribe is ε/3.)