## Finding The Stationary Values

Even without any new technique of solution, the constrained maximum in the simple example defined by (12.1) and (12.2) can easily be found. Since the constraint (12.2) implies

(12.2') x2 =---- = 30 - 2xt we can combine the constraint with the objective function by substituting (12.2') into (12.1). The result is an objective function in one variable only:

U = xt (30 - 2xt) + 2xj = 32x, - 2xf which can be handled with the method already learned. By setting dU/dx, = 32 — 4x, equal to zero, we get the solution 3c, = 8, which by virtue of (12.2') immediately leads to x2 = 30 - 2(8) = 14. From (12.1), we can then find the stationary value U = 128; and since the second derivative is d2U/dx2 = — 4 < 0, that stationary value constitutes a (constrained) maximum of U*

When the constraint is itself a complicated function, or when there are several constraints to consider, however, the technique of substitution and elimination of variables could become a burdensome task. More importantly, when the constraint comes in a form such that we cannot solve it to express one variable (x2) as an explicit function of the other (x,), the elimination method would in fact be of no avail—even if x2 were known to be an implicit function of xx, that is, even if the conditions of the implicit-function theorem were satisfied. In such cases, we may resort to a method known as the method of Lagrange (undetermined) multiplier, which, as we shall see, has distinct analytical advantages.

### Lagrange-Multiplier Method

The essence of the Lagrange-multiplier method is to convert a constrained-extremum problem into a form such that the first-order condition of the free-extremum problem can still be applied.

Given the problem of maximizing U = xtx2 + 2xt, subject to the constraint + 2x2 = 60 [from (12.1) and (12.2)], let us write what is referred to as the Lagrangian function, which is a modified version of the objective function that

* You may recall that for the flower-bed problem of Exercise 9.4-2 the same technique of substitution was applied to find the maximum area, using a constraint (the available quantity of wire netting) to eliminate one of the two variables (the length or the width of the flower bed).

incorporates the constraint as follows:

The symbol À (the Greek letter lambda), representing some as yet undetermined number, is called a Lagrange (undetermined) multiplier. If we can somehow be assured that 4xt + 2x2 = 60, so that the constraint will be satisfied, then the last term in (12.3) will vanish regardless of the value of X. In that event, Z will be identical with U. Moreover, with the constraint out of the way, we only have to seek the free maximum of Z, in lieu of the constrained maximum of U, with respect to the two variables x, and x2. The question is: How can we make the parenthetical expression in (12.3) vanish?

The tactic that will accomplish this is simply to treat À as an additional variable in (12.3), i.e., to consider Z = Z(X, xu x2). For then the first-order condition for free extremum will consist of the set of simultaneous equations

and the first equation will automatically guarantee the satisfaction of the constraint. Thus, by incorporating the constraint into the Lagrangian function Z and by treating the Lagrange multiplier as an extra variable, we can obtain the constrained extremum U (two choice variables) simply by screening the stationary values of Z, taken as a free function of three choice variables.

Solving (12.4) for the critical values of the variables, we find = 8, = 14 (and X = 4). As expected, the values of .x, and x2 check with the answers already obtained by the substitution method. Furthermore, it is clear from (12.3) that Z = 128; this is identical with the value of U found earlier, as it should be. In general, given an objective function

(12.5) z=f(x,y) subject to the constraint

(12.6) g(x,y) = c where c is a constant,* we can write the Lagrangian function as

For stationary values of Z, regarded as a function of the three variables X. x. and

* It is also possible to subsume the constant c under the contraint function so that (12.6) appears instead as G(x, y) = 0, where G(x. v) = g(x. y) - c. In that case, (12.7) should be changed to Z = f(x. y) + A[0 - G(x, y)] = f(x. y)-\G(x. r). The version in (12.6) is chosen because it facilitates the study of the comparative-static effect of a change in the constraint constant later [see (12.16)].

Since the first equation in (12.8) is simply a restatement of (12.6), the stationary values of the Lagrangian function Z will automatically satisfy the constraint of the original function z. And since the expression A[c — g(x, y)} is now assuredly zero, the stationary values of Z in (12.7) must be identical with those of (12.5), subject to (12.6).

Let us illustrate the method with two more examples.

Example 1 Find the extremum of z = xy subject to x + y = 6 The first step is to write the Lagrangian function

Z = xy + A (6 — x — y) For a stationary value of Z, it is necessary that

Thus, by Cramer's rule or some other method, we can find

The stationary value is Z = z = 9, which needs to be tested against a second-order condition before we can tell whether it is a maximum or minimum (or neither). That will be taken up later.

Example 2 Find the extremum of z = x2 + x\ subject to x, + 4x2 = 2 The Lagrangian function is

Z = x,2 + x\ 4- A(2 - x, - 4x2 ) for which the necessary condition for a stationary value is

The stationary value of Z, defined by the solution is therefore Z = z = fa. Again, a second-order condition should be consulted before we can tell whether z is a maximum or a minimum.

### Total-Differential Approach

In the discussion of the free extremum of z = f(x, y), it was learned that the first-order necessary condition may be stated in terms of the total differential dz as follows:

This statement remains valid after a constraint g(x, y ) = c is added. However, with the constraint in the picture, we can no longer take dx and dy both as "arbitrary" changes as before. For if g(x, y) = c, then dg must be equal to dc, which is zero since c is a constant. Hence.

and this relation makes dx and dy dependent on each other. The first-order necessary condition therefore becomes dz = 0 [(12.9)], subject to g = c, and hence also subject to dg = 0 [(12.10)]. By visual inspection of (12.9) and (12.10), it should be clear that, in order to satisfy this necessary condition, we must have

This result can be verified by solving (12.10) for dy and substituting the result into (12.9). The condition (12.11). together with the constraint g(x. y) = c, will provide two equations from which to find the critical values of x andy.*

Does the total-differential approach yield the same first-order condition as the Lagrange-multiplier method? Let us compare (12.8) with the result just obtained. The first equation in (12.8) merely repeats the constraint; the new result requires its satisfaction also. The last two equations in (12.8) can be rewritten, respectively, as

Zi Xy and these convey precisely the same information as (12.11). Note, however, that whereas the total-differential approach yields only the values of x and y, the Lagrange-multiplier method also gives the value of X as a direct by-product. As it turns out. A provides a measure of the sensitivity of Z (and z) to a shift of the constraint, as we shall presently demonstrate. Therefore, the Lagrange-multiplier

* Note that the constraint g = c is still to be considered along with (12.11), even though we have utilized the equation dg = 0 - that is. (12.10) —in deriving (12.11). While g = c necessarilv implies dg = 0. the converse is not true: dg = 0 merely implies g = a constant (not necessarilv c). Unless the constraint is explicitly considered, therefore, some information will be unwittingly left out of the problem.

method offers the advantage of containing certain built-in comparative-static information in the solution.

### An Interpretation of the Lagrange Multiplier

To show that X indeed measures the sensitivity of Z to changes in the constraint, let us perform a comparative-static analysis on the first-order condition (12.8). Since A, x, and y are endogenous, the only available exogenous variable is the constraint parameter c. A change in c would cause a shift of the constraint curve in the xy plane and thereby alter the optimal solution. In particular, the effect of an increase in c (a larger budget, or a larger production quota) would indicate how the optimal solution is affected by a relaxation of the constraint.

To do the comparative-static analysis, we again resort to the implicit-function theorem. Taking the three equations in (12.8) to be in the form of FJ(X, x, y; c) = 0 (with j = 1,2, 3), and assuming them to have continuous partial derivatives, we must first check that the following endogenous-variable Jacobian (where

 = gyx)
0 0