6.2 Optimization with equality constraints: n variables, m constraints
The Lagrangean for this problem is
j=1λ_{j}(g_{j}(x) − c_{j}).
As in the case of a problem with two variables and one constraint, the firstorder 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}(x_{1}, x_{2}) and g'_{2}(x_{1}, x_{2}) is nonzero—is less straightforward to generalize. The appropriate generalization involves the Jacobian matrix of the constraint functions (g_{1}, ..., g_{m}), named in honor of Carl Gustav Jacob Jacobi (1804–1851) and defined as follows.
 Definition

For j = 1, ..., m let g_{j} be a differentiable function of n variables. The Jacobian matrix of (g_{1}, ..., g_{m}) at the point x is
(∂g_{1}/∂x_{1})(x) ... (∂g_{1}/∂x_{n})(x) ... ... ... (∂g_{m}/∂x_{1})(x) ... (∂g_{m}/∂x_{n})(x) .
 Proposition 6.2.1 (Necessary conditions for an extremum)

Let f and g_{j} 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 c_{j}
for j = 1, ..., m be numbers, and suppose that x* is an interior point of S that solves the problem
max_{x} f(x) subject to g_{j}(x) = c_{j} for j = 1, ..., mor the problemmin_{x} f(x) subject to g_{j}(x) = c_{j} for j = 1, ..., mor is a local maximizer or minimizer of f(x) subject to g_{j}(x) = c_{j} for j = 1, ..., m. Suppose also that the rank of the Jacobian matrix of (g_{1}, ..., g_{m}) at the point x* is m.
Then there exist unique numbers λ_{1}, ..., λ_{m} such that x* is a stationary point of the Lagrangean function L defined by
L(x) = f(x) − ∑mThat is, x* satisfies the firstorder conditions
j=1λ_{j}(g_{j}(x) − c_{j}).L'_{i}(x*) = f_{i}'(x*) − ∑m 0 for i = 1, ..., n.
j=1λ_{j}(∂g_{j}/∂x_{i})(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

Let f and g_{j} 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
c_{j} for j = 1, ..., m be numbers. If the problem
max_{x} f(x) subject to g_{j}(x) = c_{j} for j = 1, ..., mhas a solution, it may be found as follows.
 Find all the values of (x, λ) (where λ = (λ_{1}, ..., λ_{m})) for which (a) x is an interior point of S and (b) (x, λ) satisfies the firstorder conditions given in the previous result and the constraints.
 Find all the points x for which the rank of the Jacobian matrix of (g_{1}, ..., g_{m}) is less than m and g_{j}(x) = c_{j} for j = 1, ..., m.
 If the set S has any boundary points, find all the points that solve the problem max_{x}f(x) subject to the conditions that g_{j}(x) = c_{j} 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)

Let f and g_{j} 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 g_{j}(x*) = c_{j} for j = 1, ..., m. Then
j=1λ*_{j} (g_{j}(x) − c_{j}).
 Example 6.2.1

Consider the problem
min_{x,y,z} x^{2} + y^{2} + z^{2} 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) = x^{2} + y^{2} + z^{2} − λ_{1}(x + 2y + z − 1) − λ_{2}(2x − y − 3z − 4). 1 2 1 2 −1 −3 The Lagrangean is convex for any values of λ_{1} and λ_{2}, so that any stationary point is a solution of the problem.
The firstorder conditions are
2x − λ_{1} − 2 λ_{2} = 0 2y − 2 λ_{1} + λ_{2} = 0 2z − λ_{1} + 3 λ_{2} = 0 x + 2 y + z = 1 2 x − y − 3 z = 4 λ_{1} = (2/5)x + (4/5)y λ_{2} = (4/5)x − (2/5)y. x = 16/15, y = 1/3, z = −11/15,with λ_{1} = 52/75 and λ_{2} = 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 firstorder 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 c_{j} increases.If the jth constraint arises because of a limit on the amount of some resource, we sometimes call λ_{j}(c) the shadow price of the jth resource.