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Figure 12.7 Interval constraint that changes optimal values of both variables

fx =4-2(1)+ 1.5 = 3.5 > 0. ./i(l -.<;> = () f2 = 2-2(1.5) + 1 =0. /2(1.5 — 0) = f2( 2.67

and so the conditions hold.

Figure 12.7 illustrates what is going on in this problem. The level curves rellect the concave shape of this function. The peak of the function is at (3.33. 2.67), but in the constrained problem we are restricted to the interval |(). I ] for .v,. Then the point (1. 2.671 is not on the highest attainable level curve. We reach the highest possible level curve by moving to [I. 1.5]. Note that this is a point of tungency between the vertical constraint line and the highest possible level curve.

You may be worried about the amount of guesswork in this answer and want a more systematic approach. This takes the following form. Note that, for any value of xi that may be set, a necessary condition for an optimal solution is that x2 must maximize the function for that given level of V|. Thus we can solve the problem max y = 4x| + 2X2 ~ xf - 4- X|X2 s.t. 0 < x2 < 2.67

with respect to v: taking x, as given. From the first-order condition

2 — 2x;< + x, =0 this gives the solution value for .v? as a function of Xi x2 = I + 0.5xi

Substituting forx2 in the function, we then solve max y = 4x, + 2(1 +0.5x| j

This maximum is clearly achieved at v*= 1, giving the corresponding value Xj = 1 + 0.5xf = 1.5.

Diagramatically the first step in this procedure amounts to finding the locus of points of tangency of the contours of the function with the vertical lines corresponding to each value of X|. This locus is denoted LL in figure 12.7. Then the intersection of this locus with the line drawn at X| = I gives the overall solution.

Multiplant Monopoly with Linear Costs

Suppose ¡hat a monopoly supplies its market from two plants, with cost functions:

This means that a unit of output costs \$5 to produce at plant I and \$6 at plant 2. and unit cost does not vary with output. Given the linear demand p = 100 - (0, + q2) (12.2)

the firm's profit function is

7t = KHNc/l +<•/:) - (<7i + q2)2 - 5q\ - 6q2

We wish to lind the profit-maximizing output from each plant. Applying theorem 12.1, we have

dq i

These conditions then give the equations

which, of course, have no solution. The lines defined hy the equations are parallel What went wrong?

The answer is easy to see. Plant 2's unit cost, at \$6, is always greater than plant I's unit cost, at \$5. So. it would never pay to use plant 2: we should simply set its output at zero and find the profit-maximizing output at plant I. From the above first-order condition with q2 - 0, this gives q\ - 95/2 = 47.5. It turns out that introducing nonnegativity conditions explicitly, meaning thai ¿/, > 0. i = 1. 2, resolves the problem. These of course are perfectly reasonable restrictions to impose in any case, but in this problem they are crucial. Thus we reformulate the problem as max 7r<£/i, «/a) = I00(r/| +q2) — (</i + </2)2 - ~ s.l. 0 < q, 0 < q2

This is clearly a special case of the problem considered in theorem 12.7. with a, = 0 and b, ai -foe, We iherefore need apply only (i) of the theorem. The profit-maximizing outputs q, must satisfy

TTiiql ql) = 100 - 2(9f + 9?) - 5 < 0 and </,' (100 - 2(q\ + q{) - 5) = 0 nl{ql ql)= 100 - 2(q* + q2) - 6 < 0 and qU 100 - 2(<?; + q2) - 6) = 0

There are three solution possibilities (excluding the ease where both outputs are zeroi:

0 0