Mathematical methods for economic theory

Martin J. Osborne
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6.1.1 Optimization with an equality constraint: necessary conditions for an optimum for a function of two variables

Motivation

Consider the two-variable problem
maxx,y f(x, y) subject to g(x, y) = c.
The constraint set (i.e. the set of all pairs (x, y) for which g(x, y) = c) is a set of points in (x, y) space. Suppose that it is a curve (a one-dimensional set). Assume also that the level curves of f are one-dimensional, as in the following figure. Assume that f is increasing, so that k' > k.

f(x, y) = k' f(x, y) = k g(x, y) = c

Assuming that the functions f and g are differentiable, we see from the figure that at a solution (x*, y*) of the problem, the constraint curve is tangent to a level curve of f, so that (using the equation for a tangent),

−
f'1(x*, y*)
f'2(x*, y*)
 = −
g'1(x*, y*)
g'2(x*, y*)
or
f'1(x*, y*)
g'1(x*, y*)
 = 
f'2(x*, y*)
g'2(x*, y*)
,
assuming that neither g'1(x*, y*) nor g'2(x*, y*) is zero.

Now introduce a new variable, λ, and set it equal to the common value of the quotients: λ = f'1(x*, y*)/g'1(x*, y*) = f'2(x*, y*)/g'2(x*, y*). You might think that introducing a new variable merely complicates the problem, but in fact it is a clever step that allows the condition for a maximum to be expressed in an appealing way. In addition, the variable turns out to have a very useful interpretation.

Given the definition of λ, the condition for (x*, y*) to solve the problem may be written as the two equations

f'1(x*, y*) − λg'1(x*, y*)  = 0
f'2(x*, y*) − λg'2(x*, y*)  = 0.
At a solution of the problem we need, in addition, c = g(x*, y*) (the constraint is satisfied). Thus the following conditions must be satisfied at a solution (x*, y*):
f'1(x*, y*) − λg'1(x*, y*)  = 0
f'2(x*, y*) − λg'2(x*, y*)  = 0
c − g(x*, y*)  = 0.
The first two equations can be viewed conveniently as the conditions for the derivatives of the Lagrangean
L(x, y) = f(x, y) − λ(g(x, y) − c)
with respect to x and y to be zero. They are known as the first-order conditions for the problem, and λ is called the Lagrange multiplier.

In summary, this argument suggests that if (x*, y*) solves the problem

maxx,y f(x, y) subject to g(x, y) = c
then, if neither g'1(x*, y*) nor g'2(x*, y*) is zero and (x*, y*) is in the interior of the domain, (x*, y*) is a stationary point of the Lagrangean L (and, of course, the constraint is satisfied).

This method was developed by Joseph-Louis Lagrange (1736–1813).

Necessary conditions for an optimum

Precise conditions for an optimum are given in the following result which, unlike the preceding informal argument, requires only either g'1(x*, y*) ≠ 0 or g'2(x*, y*) ≠ 0, not both of these conditions. (Recall that a continuously differentiable function is one whose partial derivatives all exist and are continuous.)
Proposition 6.1.1.1
Let f and g be functions of two variables defined on a set S that are continuously differentiable on the interior of S, let c be a number, and suppose that (x*, y*) is an interior point of S that solves the problem
maxx,y f(x, y) subject to g(x, y) = c
or the problem
minx,y f(x, y) subject to g(x, y) = c
or is a local maximizer or minimizer of f(x, y) subject to g(x, y) = c. Suppose also that either g'1(x*, y*) ≠ 0 or g'2(x*, y*) ≠ 0.

Then there is a unique number λ such that (x*, y*) is a stationary point of the Lagrangean

L(x, y) = f(x, y) − λ(g(x, y) − c).
That is, (x*, y*) satisfies the first-order conditions
L'1(x*, y*)  =  f'1(x*, y*) − λg'1(x*, y*) = 0
 L'2(x*, y*)  =  f'2(x*, y*) − λg'2(x*, y*) = 0.
In addition, g(x*, y*) = c.
Source  
For proofs, see Simon and Blume (1994), pp. 478–480, and Sydsæter (1981), Theorem 5.20 (p. 275).
We deduce from this result that the following procedure may be used to solve a maximization problem of the type we are considering.
Procedure for solving a two-variable maximization problem with an equality constraint
Let f and g be functions of two variables defined on a set S that are continuously differentiable on the interior of S, and let c be a number. If the problem maxx,yf(x, y) subject to g(x, y) = c has a solution, it may be found as follows.
  • Find all the values of (x, y, λ) in which (a) (x, y) is an interior point of S and (b) (x, y, λ) satisfies the first-order conditions and the constraint (the points (x, y, λ) for which f'1(x, y) − λg'1(x, y) = 0, f'2(x, y) − λg'2(x, y) = 0, and g(x, y) = c).
  • Find all the points (x, y) that satisfy g'1(x, y) = 0, g'2(x, y) = 0, and g(x, y) = c. (For most problems, there are no such values of (x, y). In particular, if g is linear there are no such values of (x, y).)
  • If the set S has any boundary points, find all the points that solve the problem maxx,yf(x, y) subject to the two conditions g(x, y) = c and (x, y) is a boundary point of S.
  • The points (x, y) you have found for which f(x, y) is largest are the solutions of the problem.
As before, the variant of this procedure in which the last step involves choosing the points (x, y) at which f(x, y) is smallest may be used to solve the analogous minimization problem.
Example 6.1.1.1
Consider the problem
maxx,y xy subject to x + y = 6,
where the objective and constraint functions are defined on the set of all 2-vectors, which has no boundary.

The constraint set is not bounded, so the Extreme Value Theorem does not imply that this problem has a solution.

The Lagrangean is

L(x, y) = xy − λ(x + y − 6)
so the first-order conditions are
L'1(x, y) = y − λ  =  0
 L'2(x, y) = x − λ  =  0
and the constraint is x + y − 6 = 0.

These equations have a unique solution, (x, y, λ) = (3, 3, 3). We have g'1(x, y) = 1 ≠ 0 and g'2(x, y) = 1 ≠ 0 for all (x, y), so we conclude that if the problem has a solution it is (x, y) = (3, 3).

(We can additionally argue that if the problem has a solution (x, y) then, given that x + y = 6, we have x ≥ 0 and y ≥ 0, because the value of the objective function at (3, 3) is positive and negative at any point (x, y) with x + y = 6 and either x < 0 or y < 0. Thus any solution of the problem with the additional constraints x ≥ 0 and y ≥ 0 is a solution of the problem. Now, the problem with the additional constraints has a solution because its constraint set is closed and bounded. Thus the original problem has a solution, and hence this solution is (3, 3).)

Example 6.1.1.2
Consider the problem
maxx,y x2y subject to 2x2 + y2 = 3,
where the objective and constraint functions are defined on the set of all 2-vectors, which has no boundary.

The constraint set is compact and the objective function is continuous, so the Extreme Value Theorem implies that the problem has a solution.

The Lagrangean is

L(x, y) = x2y − λ(2x2 + y2 − 3)
so the first-order conditions are
L'1(x, y)  =  2xy − 4λx = 2x(y − 2λ) = 0
 L'2(x, y)  =  x2 − 2λy = 0
and the constraint is 2x2 + y2 − 3 = 0.

To find the solutions of these three equations, first note that from the first equation we have either x = 0 or y = 2λ. We can check each possibility in turn.

  • x = 0: we have y = 31/2 and λ = 0, or y = −31/2 and λ = 0.
  • y = 2λ: we have x2 = y2 from the second equation, so either x = 1 or x = −1 from the third equation.
    • x = 1: either y = 1 and λ = 1/2, or y = −1 and λ = −1/2.
    • x = −1: either y = 1 and λ = 1/2, or y = −1 and λ = −1/2.
In summary, the first-order conditions have six solutions:
  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.
Now, g'1(x, y) = 4x and g'2(x, y) = 2y, so the only value of (x, y) for which g'1(x, y) = 0 and g'2(x, y) = 0 is (x, y) = (0, 0). At this point the constraint is not satisfied, so the only possible solutions of the problem are the solutions of the first-order conditions.

We conclude that the problem has two solutions, (x, y) = (1, 1) and (x, y) = (−1, 1).

The problem in the next example can be interpreted as the problem of a consumer with utility function xayb and budget constraint px + y = m.
Example 6.1.1.3
Consider the problem
maxx,y xayb subject to px + y = m,
where a > 0, b > 0, p > 0, and m > 0, and the objective and constraint functions are defined on the set of all points (x, y) with x ≥ 0 and y ≥ 0 (and are continuously differentiable on the interior of this set).

The Lagrangean is

L(x, y) = xayb − λ(px + y − m)
so the first-order conditions are
axa−1yb − λp  =  0
 bxayb−1 − λ  =  0

and the constraint is px + y = m. From the first two conditions we have ay = pbx. Substituting into the constraint we obtain

x = am/((a + b)p) and y = bm/(a + b),
so that (x, y) is an interior point of the domain of the objective function and
λ = [aabb/(a+b)a+b−1][ma+b−1/pa].
The value of the objective function at this point is [am/((a + b)p)]a[bm/(a + b)]b, which is positive.

We have g'1(x, y) = p and g'2(x, y) = 1, so there are no values of (x, y) for which g'1(x, y) = g'2(x, y) = 0.

The boundary of the set on which the objective function is defined is the set of points (x, y) with x = 0 or y = 0. At every such point the value of the objective function is 0.

We conclude that if the problem has a solution, it is (x, y) = (am/((a + b)p), bm/(a + b)).

The next example illustrates the possibility that the first-order conditions and constraint have no solution.
Example 6.1.1.4
Consider the problem
maxx,y x subject to x2 = 0,
where the objective and constraint functions are defined on the set of all points (x, y).

The Lagrangean is

L(x, y) = x − λx2
so the first-order conditions are
1 − 2λx  = 0
0  = 0
and the constraint is x2 = 0. From the constraint we have x = 0, which does not satisfy the first first-order condition. Thus the three equations have no solution. Now, we have g'1(x, y) = 2x and g'2(x, y) = 0. Thus we have g'1(x, y) = g'2(x, y) = 0 for (x, y) = (0, y) for any value of y. At all these points the value of the objective function is the same (namely 0). Hence if the problem has a solution, the set of solutions is the set of pairs (0, y) for all values of y.

(In fact the problem has a solution, though we cannot deduce that it does from the Extreme Value Theorem, because the constraint set, which consists of all pairs (0, y), for any value of y, is not compact.)