# Mathematical methods for economic theory

Martin J. Osborne

## 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(xy) subject to g(xy) = c.
Assume that g'2(xy) ≠ 0. By substituting for y using the constraint, we can reduce the problem to one in a single variable, x. Let h be implicitly defined by g(xh(x)) = c. Then the problem is
maxx f(xh(x)).
Define F(x) = f(xh(x)). Then
F'(x) = f'1(xh(x)) + f'2(xh(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. 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(xh(x)) = c for all x, we have

g'1(xh(x)) + g'2(xh(x))h'(x) = 0,
so that
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*) = 0 g'1(x*, y*) g'2(x*, y*) 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 0 g'1(x*, y*) g'2(x*, y*) 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(xy, λ) = f(xy) − λg(xy), 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.

In summary, we have the following result.

Proposition
Let f and g be twice differentiable functions of two variables defined on the set 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(xy) subject to the constraint g(xy) = c.
• If D(x*, y*, λ*) < 0 then (x*, y*) is a local mimimizer of f(xy) subject to the constraint g(xy) = 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
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(xy) = xy + λ(6 − x − y)
so that the determinant of the bordered Hessian of the Lagrangean is
D(xy) = 0 1 1 1 0 1 1 1 0
.
The determinant of this matrix is 1 + 1 = 2 > 0, so the point (3, 3) is indeed a local maximizer.
Example
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. (xy, λ) = (0, 31/2, 0), with f(xy) = 0.
2. (xy, λ) = (0, −31/2, 0), with f(xy) = 0.
3. (xy, λ) = (1, 1, 1/2), with f(xy) = 1.
4. (xy, λ) = (1, −1, −1/2), with f(xy) = −1.
5. (xy, λ) = (−1, 1, 1/2), with f(xy) = 1.
6. (xy, λ) = (−1, −1, −1/2), with f(xy) = −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(xy, λ) = 0 4x 2y 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, −312) is a local maximizer.