Mathematical methods for economic theory

Martin J. Osborne
Previous section Previous page Contents Text Exercises Search Next page Next section

6.1.3 Optimization with an equality constraint: sufficient conditions for a local optimum for a function of two variables

Consider the problem
maxx,y f(x, y) subject to g(x, y) = c.
Suppose that the constraint implicitly defines y as a function of x: there is a function h such that g(x, h(x)) = c for all x. Then the problem is
maxx f(x, h(x)).
Define F(x) = f(x, h(x)). Then, using the chain rule,
F'(x) = f'1(x, h(x)) + f'2(x, h(x))h'(x).
Let x* be a stationary point of F (i.e. F'(x*) = 0). A sufficient condition for x* to be a local maximizer of F is that F"(x*) < 0. Using the chain rule and Young's theorem we have
F"(x*)  =  f"11(x*, h(x*)) + 2f"12(x*, h(x*))h'(x*) + f"22(x*, h(x*))(h'(x*))2 + f'2(x*, h(x*))h"(x*).

Now, since g(x, h(x)) = c for all x, we have

g'1(x, h(x)) + g'2(x, h(x))h'(x) = 0,
so that if g'2(x, h(x)) ≠ 0 then
h'(x) = 
−g'1(x, h(x))
g'2(x, h(x))
.
Using this expression we can find h"(x*), and substitute it into the expression for F"(x*). After some manipulation, we find that
F"(x*) = 
−D(x*, y*, λ*)
(g'2(x*, y*))2
where
D(x*, y*, λ*) = 
left parenthesis 0 g'1(x*, y*) g'2(x*, y*) right parenthesis
g'1(x*, y*) f"11(x*, y*) − λ*g"11(x*, y*) f"12(x*, y*) − λ*g"12(x*, y*)
g'2(x*, y*) f"21(x*, y*) − λ*g"21(x*, y*) f"22(x*, y*) − λ*g"22(x*, y*)
and λ* is the value of the Lagrange multiplier at the solution (i.e. f'2(x*, y*)/g'2(x*, y*)).
where D(x*, y*, λ*) is the matrix
left parenthesis 0 g'1(x*, y*) g'2(x*, y*) right parenthesis
g'1(x*, y*) K"11(x*, y*, λ*) K"12(x*, y*, λ*)
g'2(x*, y*) K"21(x*, y*, λ*) K"22(x*, y*, λ*)
,
with the function K defined by K(x, y, λ) = f(x, y) − λg(x, y), and λ* is the value of the Lagrange multiplier at the solution (i.e. f'2(x*, y*)/g'2(x*, y*)).

The matrix of which D(x*, y*, λ*) is the determinant is known as the bordered Hessian of the Lagrangean.

Precisely, we can show the following result.

Proposition 6.1.3.1
Let f and g be functions of two variables defined on a set S that are twice differentiable on the interior of S, and let c be a number. Suppose that (x*, y*), an interior point of S, and the number λ* satisfy the first-order conditions
f'1(x*, y*) − λ*g'1(x*, y*)  =  0
f'2(x*, y*) − λ*g'2(x*, y*)  =  0
and the constraint g(x*, y*) = c.
  • If D(x*, y*, λ*) > 0 then (x*, y*) is a local maximizer of f(x, y) subject to the constraint g(x, y) = c.
  • If D(x*, y*, λ*) < 0 then (x*, y*) is a local minimizer of f(x, y) subject to the constraint g(x, y) = c.
Source  
For a proof, see Simon and Blume (1994), Theorem 19.7 (p. 461). (Note that although Simon and Blume's statement of the result includes the assumption that the second derivatives of f and g are continuous, this assumption is not used in their proof.)
Example 6.1.3.1
Consider the problem
maxx,yxy subject to x + y = 6.
We argued previously that if this problem has a solution, it is (3, 3). Now we can check if this point is at least a local maximizer.

The Lagrangean is

L(x, y) = xy − λ(x + y − 6)
so that the determinant of the bordered Hessian of the Lagrangean is
D(x, y, λ) =
left parenthesis 0 1 1 right parenthesis
1 0 1
1 1 0
(which is independent of x, y, and λ). The determinant of this matrix is 1 + 1 = 2 > 0, so the point (3, 3) is indeed a local maximizer.
Example 6.1.3.2
Consider the problem
maxx,y x2y subject to 2x2 + y2 = 3.
We found previously that there are six solutions of the first-order conditions, namely
  1. (x, y, λ) = (0, 31/2, 0), with f(x, y) = 0.
  2. (x, y, λ) = (0, −31/2, 0), with f(x, y) = 0.
  3. (x, y, λ) = (1, 1, 1/2), with f(x, y) = 1.
  4. (x, y, λ) = (1, −1, −1/2), with f(x, y) = −1.
  5. (x, y, λ) = (−1, 1, 1/2), with f(x, y) = 1.
  6. (x, y, λ) = (−1, −1, −1/2), with f(x, y) = −1.
Further, we found that solutions 3 and 5 are global maximizers, while solutions 4 and 6 are global minimizers.

The two remaining solutions of the first-order conditions, (0, 31/2) and (0, −31/2), are neither global maximizers nor global minimizers. Are they local maximizers or local minimizers?

The determinant of the bordered Hessian of the Lagrangean is

D(x, y, λ) = 
left parenthesis 0 4x 2y right parenthesis
4x 2y−4λ 2x
2y 2x −2λ
.
This determinant is
−4x(−8xλ − 4xy) + 2y(8x2 − 2y(2y − 4λ)),
which simplifies to
8[2λ(2x2 + y2) + y(4x2 − y2)].
Given that 2x2 + y2 = 3 at each solution, from the constraint, the determinant is thus
8[6λ + y(4x2 − y2)].
The value of this determinant at the two solutions is
  • (0, 31/2, 0): −8·33/2, so (0, 31/2) is a local minimizer;
  • (0, −31/2, 0): 8·31/2, so (0, −31/2) is a local maximizer.