Analysis

6.5 The Method Of Lagrange Multipliers

As an application of the implicit function theorem, consider the method of Lagrange multipliers from calculus. Recall the problem is to maximize or minimize a function subject to equality constraints. Let f : U → ℝ be a C¹ function where U ⊆ ℝⁿ and let

gi(x) = 0, i = 1,⋅⋅⋅,m (6.15)

(6.15)

be a collection of equality constraints with m < n. Now consider the system of nonlinear equations

x₀ is a local maximum if f ≥ f for all x near x₀ which also satisfies the constraints 6.15. A local minimum is defined similarly. Let F : U × ℝ → ℝ^m+1 be defined by

( f (x)− a ) | g1(x) | F (x,a) ≡ || . || . (6.16) ( .. ) gm (x )

(6.16)

Now consider the m + 1 × n Jacobian matrix, the matrix of the linear transformation, D₁F

with respect to the usual basis for ℝⁿ and ℝ^m+1.

( f (x ) ⋅⋅⋅ f (x ) ) | gx1(x0) ⋅⋅⋅ gxn (x0) | || 1x1. 0 1xn. 0 || . ( .. .. ) gmx1 (x0) ⋅⋅⋅ gmxn (x0)

If this matrix has rank m + 1 then some m + 1 ×m + 1 submatrix has nonzero determinant. It follows from the implicit function theorem that there exist m + 1 variables, x_i1,

,x_im+1 such that the system

F (x,a) = 0 (6.17)

(6.17)

specifies these m + 1 variables as a function of the remaining n−

variables and a in an open set of ℝ^n−m. Thus there is a solution to 6.17 for some x close to x₀ whenever a is in some open interval. Therefore, x₀ cannot be either a local minimum or a local maximum. It follows that if x₀ is either a local maximum or a local minimum, then the above matrix must have rank less than m + 1 which requires the rows to be linearly dependent. Thus, there exist m scalars,

λ1,⋅⋅⋅,λm,

and a scalar μ, not all zero such that

( ) ( ) ( ) fx1 (x0) g1x1 (x0 ) gmx1 (x0) μ |( .. |) = λ1|( .. |) + ⋅⋅⋅+ λm |( .. |) . (6.18) . . . fxn (x0) g1xn (x0) gmxn (x0)

(6.18)

If the column vectors

( g1x (x0) ) ( gmx (x0) ) | 1. | | 1. | ( .. ) ,⋅⋅⋅( .. ) (6.19) g1xn (x0) gmxn (x0)

(6.19)

are linearly independent, then, μ≠0 and dividing by μ yields an expression of the form

( ) ( ) ( ) fx1 (x0) g1x1 (x0) gmx1 (x0) |( ... |) = λ1|( ... |) + ⋅⋅⋅+ λm |( ... |) (6.20) fx (x0) g1x (x0) gmx (x0) n n n

(6.20)

at every point x₀ which is either a local maximum or a local minimum. This proves the following theorem.