6.2 Optimization with equality constraints: n variables, m constraints
The Lagrangean for this problem is
j=1λj(gj(x) − cj).
As in the case of a problem with two variables and one constraint, the first-order condition is that x* be a stationary point of the Lagrangean. The “nondegeneracy” condition in the two variable case—namely that at least one of g'1(x1, x2) and g'2(x1, x2) is nonzero—is less straightforward to generalize. The appropriate generalization involves the Jacobian matrix of the constraint functions (g1, ..., gm), named in honor of Carl Gustav Jacob Jacobi (1804–1851) and defined as follows.
- Definition
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For j = 1, ..., m let gj be a differentiable function of n variables. The Jacobian matrix of (g1, ..., gm) at the point x is
(∂g1/∂x1)(x) ... (∂g1/∂xn)(x) ... ... ... (∂gm/∂x1)(x) ... (∂gm/∂xn)(x) .
- Proposition 6.2.1 (Necessary conditions for an extremum)
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Let f and gj for j = 1, ..., m be functions of n variables defined on a set S, with m ≤ n, that are continuously differentiable on the interior of S, let cj
for j = 1, ..., m be numbers, and suppose that x* is an interior point of S that solves the problem
maxx f(x) subject to gj(x) = cj for j = 1, ..., mor the problemminx f(x) subject to gj(x) = cj for j = 1, ..., mor is a local maximizer or minimizer of f(x) subject to gj(x) = cj for j = 1, ..., m. Suppose also that the rank of the Jacobian matrix of (g1, ..., gm) at the point x* is m.
Then there exist unique numbers λ1, ..., λm such that x* is a stationary point of the Lagrangean function L defined by
L(x) = f(x) − ∑mThat is, x* satisfies the first-order conditions
j=1λj(gj(x) − cj).L'i(x*) = fi'(x*) − ∑m 0 for i = 1, ..., n.
j=1λj(∂gj/∂xi)(x*) =
- Source
- For proofs, see Sydsæter (1981), Theorem 5.20 (p. 275) and Simon and Blume (1994), pp. 478–480. (Only Sydsæter argues explicitly that the Lagrange multipliers are unique.)
- Procedure for solving a maximization problem with many equality constraints
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Let f and gj for j = 1, ..., m be functions of n variables defined on a set S, with m ≤ n, that are continuously differentiable on the interior of S, and let
cj for j = 1, ..., m be numbers. If the problem
maxx f(x) subject to gj(x) = cj for j = 1, ..., mhas a solution, it may be found as follows.
- Find all the values of (x, λ) (where λ = (λ1, ..., λm)) for which (a) x is an interior point of S and (b) (x, λ) satisfies the first-order conditions given in the previous result and the constraints.
- Find all the points x for which the rank of the Jacobian matrix of (g1, ..., gm) is less than m and gj(x) = cj for j = 1, ..., m.
- If the set S has any boundary points, find all the points that solve the problem maxxf(x) subject to the conditions that gj(x) = cj for j = 1, ..., m and x is a boundary point of S.
- The points x you have found for which f(x) is largest are the solutions of the problem.
- Proposition 6.2.2 (Conditions under which necessary conditions for an extremum are sufficient)
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Let f and gj for j = 1, ..., m be functions of n variables defined on a convex set S that are continuously differentiable on the interior of S. Suppose that there are numbers
λ*1, ..., λ*m and an interior point x* of S such that x* is a stationary point of the Lagrangean
L(x) = f(x) − ∑mSuppose further that gj(x*) = cj for j = 1, ..., m. Then
j=1λ*j (gj(x) − cj).
- Example 6.2.1
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Consider the problem
minx,y,z x2 + y2 + z2 subject to x + 2y + z = 1 and 2x − y − 3z = 4with the domain of the objective function the set of all triples (x, y, z) of real numbers, which has no boundaries.
The Lagrangean is
L(x, y, z) = x2 + y2 + z2 − λ1(x + 2y + z − 1) − λ2(2x − y − 3z − 4). 1 2 1 2 −1 −3 The Lagrangean is convex for any values of λ1 and λ2, so that any stationary point is a solution of the problem.
The first-order conditions are
2x − λ1 − 2 λ2 = 0 2y − 2 λ1 + λ2 = 0 2z − λ1 + 3 λ2 = 0 x + 2 y + z = 1 2 x − y − 3 z = 4 λ1 = (2/5)x + (4/5)y λ2 = (4/5)x − (2/5)y. x = 16/15, y = 1/3, z = −11/15,with λ1 = 52/75 and λ2 = 54/75.We conclude that (x, y, z) = (16/15, 1/3, −11/15) is the unique solution of the problem.
Interpretation of Lagrange multipliers
In the case of a problem with two variables and one constraint we saw that the Lagrange multiplier is equal to the rate of increase of the maximal value of the objective function as the constraint is relaxed. This result has a natural generalization to problems with n variables and m constraints.Consider the problem
Let
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 for the jth constraint at the solution is equal to the rate of change in the maximal value of the objective function as cj increases.If the jth constraint arises because of a limit on the amount of some resource, we sometimes call λj(c) the shadow price of the jth resource.