Optimization over an Interval

The discussion of first- and second-order conditions in sections 6 I and 6.2 has dealt exclusively with the unconstrained case, in which a solution to the problem can be anywhere on the real line. Often in economics, however, this is unacceptably general. For example, in problems in which firms choose outputs (as in several examples in the previous section ) or consumers choose goods, we cannot assume that negative quantities are possible. In other problems, it may be reasonable to place an upper bound on a variable: for example, a firm may have a fixed production capacity that puts an upper limit on how much output it can produce. Or alternatively, a decision-taker may be choosing a proportion, for example, the share of a company to buy, and lhat naturally places the bounds zero and one on the variable.

Example 6.12 Solve the following problems involving optimization over an interval:

(iv) min y = 2xi - 0.5jc2 + 2 subject to 0 < .t < I tv) miny = 4.v2 - 5.x + ¡0subject to ( < .v < 10


(i) This is a linear function with a negative slope. Therefore, we know an interior maximum cannot occur, and the solution is clearly at x1 = 0. with y* = 3. Note that at x = 0 (and indeed at all points in the domain)

(ii) This is a constant function. All the points in the interval [-1. 1] are solutions because they all yield the maximum and minimum of the function (see figure 6.25).

Figure 6.24 Graph oí the problem in example 6.12(ii

Figure 6.25 Graph of thc problein in example 6.12(ii)

(iii) This function takes on a minimum at a " - 0. Thus we have an interior solution (see figure 6.26).

(iv) We solved this problem first for unrestricted x as (i) of examples 6.1 and 6.7. The unconstrained minimum of the function occurs at a* = 0.167. This value of x is not ruled out by the constraint. Hence we have an interior solution with a* = 0.167. Note that f(x') = 1.995, which is less than /(0) - 2. so v" does indeed deliver the minimum value on this interval (see figure 6.27).

,, v = 6r
0 0

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