SM286A – Mathematics for Economics
Asst. Prof. Nelson Uhan
Fall 2015
Lesson 25. Optimization with Equality Constraints, cont.
1
Overview
● Last time: the Lagrange multiplier method for optimization problems with one equality constraint
● Today: multiple equality constraints
2 The Lagrange multiplier method – m equality constraints
f (x1 , . . . , x n )
minimize/maximize
g1 (x1 , . . . , x n ) = c1
subject to
⋮
g m (x1 , . . . , x n ) = c m
● Step 1. Introduce Lagrange multipliers λ1 , . . . , λ m for each equality constraint and form the Lagrangian
function Z:
Z(x1 , . . . , x n , λ1 , . . . , λ m ) = f (x1 , . . . , x n ) + λ1 [c1 − g1 (x1 , . . . , x n )] + ⋅ ⋅ ⋅ + λ m [c m − g m (x1 , . . . , x n )]
● Step 2. Find the critical points by solving the system of equations implied by the first-order necessary
condition:
∇Z(x1 , . . . , x n , λ1 , . . . , λ m ) = 0
● Step 3. Classify each critical point as a local minimum or local maximum by applying the second-order
sufficient condition:
○ The bordered Hessian matrix H is
⎡ 0
⎢
⎢
⎢
⎢ 0
⎢
⎢ ⋮
⎢
⎢
⎢ 0
⎢
H = ⎢⎢ ∂g1
⎢ ∂x
⎢ 1
⎢ ∂g1
⎢
⎢ ∂x2
⎢
⎢ ⋮
⎢
⎢ ∂g1
⎢ ∂x
⎣ n
0
...
0
0
⋮
0
...
⋮
...
0
⋮
0
∂g 2
∂x 1
∂g 2
∂x 2
...
∂g m
∂x 1
∂g m
∂x 2
⋮
⋮
...
∂g 2
∂x n
...
∂g 1
∂x 1
∂g 2
∂x 1
∂g 1
∂x 2
∂g 2
∂x 2
...
...
⋮
...
⋮
⋮
∂g m
∂x 1
∂g m
∂x 2
∂2 Z
∂x 12
∂2 Z
∂x 2 ∂x 1
∂2 Z
∂x 1 ∂x 2
∂2 Z
∂x 22
...
⋮
...
⋮
⋮
⋮
∂g m
∂x n
∂2 Z
∂x n ∂x 1
∂2 Z
∂x n ∂x 2
...
∂g 1
∂x n
∂g 2
∂x n
⋮
∂g m
∂x n
∂2 Z
∂x 1 x n
∂2 Z
∂x 2 x n
⋮
∂2 Z
∂x n2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
○ The ith bordered leading principal minor of H — denoted by ∣H i ∣ — is the determinant of the
square submatrix formed by the first m + i rows and columns of H
1
○ Let (a1 , . . . , a n ) be a critical point found in Step 2. Then
(i) f (a1 , a2 , . . . , a n ) is a local minimum if
∣H m+1 ∣, ∣H m+2 ∣, . . . , ∣H n ∣ all have the same sign as (−1)m
(ii) f (a1 , a2 , . . . , a n ) is a local maximum if
∣H m+1 ∣ has the same sign as (−1)m+1
and
∣H m+1 ∣, ∣H m+2 ∣, . . . , ∣H n ∣ alternate in sign
Example 1. Use the Lagrange multiplier method to find the local optima of
minimize/maximize
subject to
x3
x1 + x2 + x3 = 12
x12 + x22 − x3 = 0
● In this problem, n =
and m =
Step 1. Introduce the Lagrange multipliers λ1 , . . . , λ m and form the Lagrangian function Z.
● The Lagrangian function Z is
Step 2. Find the critical points.
● The gradient of Z is
● The first-order necessary condition tells us that the critical points must satisfy
2
● Solving this system of equations, we find that there are two critical points:
Step 3. Classify the critical points as a local minimum or local maximum.
● The bordered Hessian is
● The bordered Hessian at the critical point (x1 , x2 , x3 , λ1 , λ2 ) = (2, 2, 8, 4/5, −1/5) is
● The bordered leading principal minors ∣H m+1 ∣, ∣H m+2 ∣, . . . of H(2, 2, 8, 4/5, −1/5) are
● Therefore,
3
● The bordered Hessian at the critical point (x1 , x2 , x3 , λ1 , λ2 ) = (−3, −3, 18, 6/5, 1/5) is
● The bordered leading principal minors ∣H m+1 ∣, ∣H m+2 ∣, . . . of H(−3, −3, 18, 6/5, 1/5) are
● Therefore,
Example 2. Use the Lagrange multiplier method to find the local optima of
minimize/maximize
subject to
x12 + x22 + x32
3x1 + x2 + x3 = 5
x1 + x2 + x3 = 1
4