Mathematical methods for economic theory

Martin J. Osborne

6.1.4 Optimization with an equality constraint: conditions under which a stationary point is a global optimum

We know that if (x*, y*) solves the problem
maxx,y f(xy) subject to g(xy) = c
and either g'1(x*, y*) ≠ 0 or g'2(x*, y*) ≠ 0 then there is a number λ* such that (x*, y*) is a stationary point of the Lagrangean L(xy) = f(xy) − λ*(g(xy) − c)), given λ*.

The fact that (x*, y*) is a stationary point of the Lagrangean does not mean that (x*, y*) maximizes the Lagrangean, given λ*. (The Lagrangean is a function like any other, and we know that a stationary point of an arbitrary function is not necessarily a maximizer of the function. In an exercise you are asked to work through a specific example.)

Suppose, however, that (x*, y*) does in fact maximize L(xy), given λ*. Then

L(x*, y*) ≥ L(xy) for all (xy),
or
f(x*, y*) − λ*(g(x*, y*) − c) ≥ f(xy) − λ*(g(xy) − c) for all (xy).
Now, if (x*, y*) satisfies the constraint then g(x*, y*) = c, so this inequality is equivalent to
f(x*, y*) ≥ f(xy) − λ*(g(xy) − c) for all (xy),
so that
f(x*, y*) ≥ f(xy) for all (xy) with g(xy) = c.
That is, (x*, y*) solves the constrained maximization problem.

In summary,

if (x*, y*) maximizes the Lagrangean, given λ*, and satisfies the constraint, then it solves the problem.
Now, we know that any stationary point of a concave function is a maximizer of the function. Thus if the Lagrangean is concave in (xy), given λ*, and (x*, y*) is a stationary point of the Lagrangean, then (x*, y*) maximizes the Lagrangean, given λ*, and hence if it satisfies the constraint then it solves the problem. Precisely, we have the following result.
Proposition  
Suppose that f and g are differentiable functions of two variables defined on an open convex set S and suppose that there exists a number λ* and an interior point (x*, y*) of S that is a stationary point of the Lagrangean
L(xy) = f(xy) − λ*(g(xy) − c).
That is,
L'1(x*, y*)  =  f'1(x*, y*) − λ*g'1(x*, y*) = 0
L'2(x*, y*)  =  f'2(x*, y*) − λ*g'2(x*, y*) = 0.
Suppose further that g(x*, y*) = c. Then
  • if L is concave—in particular if f is concave and λ*g is convex—then (x*, y*) solves the problem maxx,y f(xy) subject to g(xy) = c
  • if L is convex—in particular if f is convex and λ*g is concave—then (x*, y*) solves the problem minx,y f(xy) subject to g(xy) = c.
Proof  
Given that f and g are differentiable, so is L, so the fact that L is concave means that by a previous result any stationary point of L is a maximizer of L. The argument before the statement of the result establishes that any maximizer of L that satisfies the constraint g(xy) = c solves the constrained maximization problem.
If g is linear then λ*g is both convex and concave, regardless of the value of λ*. Thus this result has the following corollary.
Corollary
Suppose that f is a continuously differentiable function of two variables defined on an open convex set S and g is a linear function of two variables defined on this set. Suppose that there exists a number λ* and an interior point (x*, y*) of S that is a stationary point of the Lagrangean
L(xy) = f(xy) − λ*(g(xy) − c).
Suppose further that g(x*, y*) = c. Then
  • if f is concave then (x*, y*) solves the problem maxx,y f(xy) subject to g(xy) = c
  • if f is convex then (x*, y*) solves the problem minx,y f(xy) subject to g(xy) = c.
Example
Consider the problem
maxx,y xayb subject to px + y = m,
considered previously. We found that there is a value of λ* such that
(x*, y*) =  left parenthesis
am
(a + b)p
bm
a + b
right parenthesis
is a stationary point of the Lagrangean and satisfies the constraint. Now, if a ≥ 0, b ≥ 0, and a + b ≤ 1 then the objective function is concave; the constraint is linear, so from the result above, (x*, y*) is a solution of the problem.