## Info

Figure 6.28 Graph of ihe problem in example 6.12(v)

Fori < x < 10. this slope is always positive. Thus we know that the minimum vitlue of the function over this interval occurs at a = I and the maximum at .v = 10 (see figure 6.28). ■

### Competitive Firm with Linear Costs Revisited

Suppose that a firm has the total-cost function C - 2().v and sells into a competitive market where Lhe given price is $10 per unit. Its profit function is jr(jr) = lO.t — 20.x = -10.v and the first-order condition gives jt'(.v* ) = —10 = 0

which, of course, cannot be satisfied. The intuitive answer to the problem is obvious: market price is below unit cost (horizontal demand curve below horizontal marginal-cost curve), and so the firm does best by producing zero output. However, this possibility is not captured in the mathematics because, implicitly, we have allowed the whole real line. The mathematical solution is to set output at —oo, because multiplying this by —10 gives the largest possible profit! If it is the case that negative outputs are impossible, we should incorporate this into the problem explicitly by specifying the constraint

Then, as we will show below, the mathematics will produce the correct answer.

Returning to ihe general case, we assume that the possible values of x that are feasible for the problem are determined by a constraint a < x < b which delines an interval on the real line. We then write the problem as max f(x) subject to a<x<b (6.10)

We are now trying to find the highest value of the function over this given interval. Figure 6.29 illustrates this situation. We see immediately that it is no longer a necessary condition for an extreme value that the derivative is zero.

Figure 6.29 (a) Local constrained optima with interval constraint binding (b) Local constrained optima with interval constraint not binding

The function f shown in figure 6.29 (a) has a local (and global) maximum at b and a local (and global) minimum at a, while fib) > 0, /'(«) > 0. The reason b gives a local maximum, although f'(b) > 0. is that we cannot increase -y above b: the only feasible direction of change in x is to go below b, and this reduces J . Likewise, a is a local minimum because the only feasible change in .r increases the value of the junction. On the other hand, figure 6.29 (b) also shows an example of a function g which has a maximum at a *, and the standard condition g'(jr*) = 0 applies.

We first develop first-order necessary conditions for the maximization problem in equation (6.10). Let a * denote a local maximum for this problem (so that necessarily a < x* £ b). There are then three possibilities:

I. .Y* = o. In this case we must have /'(.x*) < 0. To see this, consider the differential dy = f'(u) dx

When x = a, the only permissible change in x is dx > 0, and so to make dy < 0, we must have /'(a) < 0.

2. a < x* < b. In this case we must have /'(x*) = 0. This is because, when x* is inside the interval, we can choose dx to be both positive or negative, and so for dy not to be positive we must have f'(x*) = 0.

3. x* = b. In this case we must have fix") > 0. Again, consider the differential dy = f'(b)dx

When x = b. the only permissible change in x is dx < 0. For this not to produce dy > 0, we must have fib) > 0.

These arguments simply formalize somewhat the discussion of figures 6.28 and 6.29 and the results can be stated succinctly as follows:

Theorem 6,4 |
If .v* is a solution to the problem |

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