6.1.4 Optimization with an equality constraint: conditions under which a stationary point is a global optimum
maxx,y f(x, y) subject to g(x, y) = c,
the concavity of the Lagrangean
L(x, y) = f(x, y) − λ(g(x, y) − c))
guarantees that a stationary point of the Lagrangean is a maximizer.
The argument is simple. If L is concave, given λ, then by a previous result any stationary point of L, given λ, is a maximizer of L. That is, if (x*, y*) is a stationary point of L then
L(x*, y*) ≥ L(x, y) for all (x, y),
or
f(x*, y*) − λ(g(x*, y*) − c) ≥ f(x, y) − λ(g(x, y) − c) for all (x, y).
Now, if (x*, y*) satisfies the constraint then g(x*, y*) = c, this inequality is equivalent to
f(x*, y*) ≥ f(x, y) − λ(g(x, y) − c) for all (x, y),
so that
f(x*, y*) ≥ f(x, y) for all (x, y) with g(x, y) = c.
That is, (x*, y*) solves the constrained maximization problem.
Precisely, we have the following result.
- Proposition 6.1.4.1
-
Suppose that f and g are functions of two variables defined on a convex set S that are differentiable on the interior of S, c is a number, and there exists a number λ* and an interior point
(x*, y*) of S such that (x*, y*) is a stationary point of the Lagrangean
L(x, y) = f(x, y) − λ*(g(x, y) − 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.
- 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(x, y) = c solves the constrained maximization problem.
- Corollary 6.1.4.1
-
Suppose that f is a function of two variables defined on a convex set S that is differentiable on the interior of S, g is a linear function of two variables defined on S, c is a number, and there exists a number λ*
and an interior point (x*, y*) of S such that (x*, y*) is a stationary point of the Lagrangean
L(x, y) = f(x, y) − λ*(g(x, y) − c).Suppose also that g(x*, y*) = c.
- Example 6.1.4.1
-
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*) = am (a + b)p , bm a + b