Mathematical methods for economic theory

Martin J. Osborne

5.1 Necessary conditions for an interior optimum

One variable

From your previous study of mathematics, you probably know that if the function f of a single variable is differentiable and I is an interval of numbers then there is a relationship between the solutions of the problem
maxxf(x) subject to x ∈ I
and the points in I at which the first derivative of f is zero—that is, the stationary points of f.
Definition
Let f be a function of a single variable defined on a set S. A point x ∈ S at which f is differentiable and f'(x) = 0 is a stationary point of f.
What precisely is this relationship? Consider the cases in the three figures.

a b x* f(x)    a b x* f(x) x' x''    a b x' f(x) x''

  • The function in the left figure has a unique stationary point x*, which is the global maximizer of the function.
  • The function in the middle figure has three stationary points, x*, x', and x". The point x* is the global maximizer of the function, while x' is a local (though not global) minimizer and x" is a local (but not global) maximizer.
  • The function in the right figure has two stationary points, x' and x". The point x' in neither a local maximizer nor a local minimizer; x" is a global minimizer.
We see that
  • a stationary point is not necessarily a global maximizer, or even a local maximizer, or even a local optimizer of any sort (maximizer or minimizer) (consider x' in the right-hand figure)
  • a global maximizer is not necessarily a stationary point (consider a in the right-hand figure).
That is, being a stationary point is neither a necessary condition nor a sufficient condition for solving the problem. So what is the relation between stationary points and maximizers?

Although a maximizer may not be a stationary point, the only case in which it is not is when it is one of the endpoints of the interval I on which f is defined. That is, any point interior to this interval that is a maximizer must be a stationary point.

Proposition  proof
Let f be a function of a single variable defined on the interval I. If a point x in the interior of I is a local maximizer or minimizer of f and f is differentiable at x then f'(x) = 0.
Proof  hide
Suppose that x is a local maximizer of f. Because x is in the interior of I, for h > 0 sufficiently small we have x + h ∈ I, so that f(x + h) is defined. Thus because x is a local maximizer of f, for small enough values of h we have f(x + h) ≤ f(x), and hence (f(x + h) − f(x))/h ≤ 0. The limit of left-hand side of this inequality as h → 0 is f'(x) (by the definition of a derivative), so that f'(x) ≤ 0. A symmetric argument using h < 0 shows that f'(x) ≥ 0. Thus f'(x) = 0. A similar argument applies when x is a local minimizer of f.
This result gives a necessary condition for x to be a maximizer (or a minimizer) of f: if x is a maximizer (or a minimizer) of f, x is not an endpoint of I, and f is differentiable at x, then x is a stationary point of f. The condition is obvously not sufficient for a point to be a maximizer—the condition is satisfied also, for example, at points that are minimizers. The first-derivative is involved, so we refer to the condition as a first-order condition.

Thus among all the points in the interval I, only the endpoints (if any) and the stationary points of f can be maximizers of f. Most functions have a relatively small number of stationary points, so the following procedure to find the maximizers is useful.

Procedure for solving a single-variable maximization problem on an interval
Let f be a differentiable function of a single variable and let I be an interval. If the problem maxxf(x) subject to x ∈ I has a solution, it may be found as follows.
  • Find all the stationary points of f (the points x for which f'(x) = 0) that are in I, and calculate the values of f at each such point.
  • Find the values of f at the endpoints, if any, of I.
  • The points x you have found at which the value f(x) is largest are the maximizers of f.
The variant of this procedure in which the last step involves choosing the points x at which f(x) is smallest may be used to solve the analogous minimization problem.
Example
Consider the problem
maxx x2 subject to x ∈ [−1, 2].
This problem satisfies the conditions of the extreme value theorem, and hence has a solution. Let f(x) = x2. We have f'(x) = 2x, so the function has a single stationary point, x = 0, which is in the constraint set. The value of the function at this point is f(0) = 0. The values of f at the endpoints of the interval on which it is defined are f(−1) = 1 and f(2) = 4. Thus the global maximizer of the function on [−1, 2] is x = 2 and the global minimizer is x = 0.
Example
Consider the problem
maxx −x2 subject to x ∈ (−∞, ∞).
This problem does not satisfy the conditions of the extreme value theorem, so that the theorem does not tell us whether it has a solution. Let f(x) = −x2. We have f'(x) = −2x, so that the function has a single stationary point, x = 0, which is in the constraint set. The constraint set has no endpoints, so x = 0 is the only candidate for a solution to the problem. We conclude that if the problem has a solution then the solution is x = 0. In fact, the problem does have a solution: we have f(x) ≤ 0 for all x and f(0) = 0, so the solution is indeed x = 0.
Example
Consider the problem
maxx x2 subject to x ∈ (−∞, ∞).
Like the problem in the previous example, this problem does not satisfy the conditions of the extreme value theorem, so that the theorem does not tell us whether it has a solution. Let f(x) = x2. We have f'(x) = 2x, so that the function has a single stationary point, x = 0, which is in the constraint set. The constraint set has no endpoints, so x = 0 is the only candidate for a solution to the problem. We conclude that if the problem has a solution then the solution is x = 0. In fact, the problem does not have a solution: the function f increases without bound as x increases (or decreases) without bound.

Many variables

Consider a maximum of a function of two variables. At this maximum the function must decrease in every direction (otherwise the point would not be a maximum!). In particular, the maximum must be a maximum along a line parallel to the x-axis and also a maximum along a line parallel to the y-axis. Hence, given the result for a function of a single variable, at the maximum both the partial derivative with respect to x and the partial derivative with respect to y must be zero. Extending this idea to many dimensions gives us the following result, where f'i is the partial derivative of f with respect to its ith argument.
Proposition  proof
Let f be a function of n variables defined on the set S. If the point x in the interior of S is a local maximizer or minimizer of f and the partial derivative of f with respect to its ith argument exists at x then
f'i(x) = 0.
In particular, if all of the partial derivatives of f exist at x then
f'j(x) = 0 for j = 1, ..., n.
Proof  hide
Suppose that x is a local maximizer of f. Because x is in the interior of S, for some r > 0 the set of all n-vectors h with |h| < r is a subset of S. Define the function g of a single variable by g(zi) = f(x1, ..., xi−1zixi+1, ..., xn) for zi ∈ (xi − rxi + r). The fact that the partial derivative of f with respect to its ith argument exists at x means that g is differentiable at xi, by the definition of a partial derivative, and g'(xi) = f'i(x). Also, given that x is a local maximizer of f, xi is a local maximizer of g. Thus by a previous result, g'(xi) = 0. Thus f'i(x) = 0. A similar argument applies when x is a local minimizer of f.
As for the analogous result for functions of a single variable, this result gives a necessary condition for a maximum (or minimum): if a point is a maximizer then it satisfies the condition. As before, the conditions are called first-order conditions and a point that satisfies them is a “stationary point”.
Definition
Let f be a function of n variables defined on a set S. A point x ∈ S at which f is differentiable and f'i(x) = 0 for i = 1, ..., n is a stationary point of f.
As for functions of a single variable, the result tells us that the only points that can be global maximizers are either stationary points or boundary points of the set S. Thus the following procedure locates all global maximizers and global minimizers of a differentiable function.
Procedure for solving a many-variable maximization problem on a set
Let f be a differentiable function of n variables and let S be a set of n-vectors. If the problem maxxf(x) subject to x ∈ S has a solution, it may be found as follows.
  • Find all the stationary points of f (the points x for which f'i(x) = 0 for i = 1, ..., n) in S and calculate the value of f at each point.
  • Find the largest and smallest values of f on the boundary of S.
  • The points x you have found at which the value of f is largest are the maximizers of f.
This method is much less generally useful than the analogous method for functions of a single variable because for many problems finding the largest and smallest values of f on the boundary of S is difficult. For this reason, we devote considerable attention to other, better methods for finding maxima and minima of maximization problems with constraints, in the next two parts.

Here are some examples, however, where the method may be fairly easily applied.

Example
Consider the problem
maxx,y[−(x − 1)2 − (y + 2)2] subject to −∞ < x < ∞ and −∞ < y < ∞.
This problem does not satisfy the conditions of the extreme value theorem (because the constraint set is not bounded), so the theorem does not tell us whether the problem has a solution. The first-order conditions are
−2(x − 1)  = 0
−2(y + 2)  = 0,
which have a unique solution, (xy) = (1, −2). The constraint set has no boundary points, so we conclude that if the problem has a solution, this solution is (xy) = (1, −2). In fact, the problem does have a solution, because the value of the objective function at (1, −2) is 0, and its value at any point is nonpositive.
Example
Consider the problem
maxx,y[(x − 1)2 + (y − 1)2] subject to 0 ≤ x ≤ 2 and −1 ≤ y ≤ 3.
This problem satisfies the conditions of the extreme value theorem, and hence has a solution. The first-order conditions are
2(x − 1)  = 0
2(y − 1)  = 0,
which have a unique solution, (xy) = (1, 1), which is in the constraint set. The value of the objective function at this point is 0.

Now consider the behavior of the objective function on the boundary of the constraint set, which is a rectangle.

  • If x = 0 and −1 ≤ y ≤ 3 then the value of the objective function is 1 + (y − 1)2. The problem of finding y to maximize this function subject to −1 ≤ y ≤ 3 satisfies the conditions of the extreme value theorem, and thus has a solution. The first-order condition is 2(y − 1) = 0, which has a unique solution y = 1, which is in the constraint set. The value of the objective function at this point is 1. On the boundary of the set {(0, y): −1 ≤ y ≤ 3}—namely at the points (0, −1) and (0, 3)—the value of the objective function is 5. Thus on this part of the boundary, the points (0, −1) and (0, 3) are the only candidates for a solution of the original problem.
  • A similar analysis leads to the conclusion that the points (2, −1) and (2, 3) are the only candidates for a maximizer on the part of the boundary for which x = 2 and −1 ≤ y ≤ 3, the points (0, −1) and (2, −1) are the only candidates for a maximizer on the part of the boundary for which 0 ≤ x ≤ 2 and y = −1, and the points (0, 3) and (2, 3) are the only candidates for a maximizer on the part of the boundary for which 0 ≤ x ≤ 2 and y = 3.
  • The value of the objective function at all these candidates for a solution on the boundary of the constraint set is 5.
Finally, comparing the values of the objective function at the candidates for a solution that are (a) interior to the constraint set (namely (1, 1)) and (b) on the boundary of the constraint set, we conclude that the problem has four solutions, (0, −1), (0, 3), (2, −1), and (2, 3).
Example
Consider the problems
maxx,y x2 + y2 + y − 1 subject to x2 + y2 ≤ 1
and
minx,y x2 + y2 + y − 1 subject to x2 + y2 ≤ 1.
In each case the constraint set, {(xy): x2 + y2 ≤ 1}, is compact. The objective function is continuous, so by the extreme value theorem, the problem has a solution.

We apply the procedure as follows, denoting the objective function by f.

  • We have f'1(xy) = 2x and f'2(xy) = 2y + 1, so the stationary points are the solutions of 2x = 0 and 2y + 1 = 0. Thus the function has a single stationary point, (xy) = (0, −1/2), which is in the constraint set. The value of the function at this point is f(0, −1/2) = −5/4.
  • The boundary of the constraint set is the set of points (xy) such that x2 + y2 = 1, as shown in the following figure.

    x y 0 1 −1 1 −1 Boundary of constraint set

    Thus for a point (xy) on the boundary we have f(xy) = x2 + 1 − x2 + y − 1 = y. We have −1 ≤ y ≤ 1 on the boundary, so the maximum of the function on the boundary is 1, which is achieved at (xy) = (0, 1), and the minimum is −1, achieved at (xy) = (0, −1).

  • Looking at all the values we have found, we see that the global maximum of f is 1, achieved at (0, 1), and the global minimum is −5/4, achieved at (0, −1/2).
Notice that the reasoning about the behavior of the function on the boundary of the constraint set is straightforward in this example because we are able easily to express the value of the function on the boundary in terms of a single variable (y). In many other problems, doing so is not easy.