Basic mathematics of economic models. 3. Maximization

Transcription

1 John Riley 1 January 16 Basic mathematics o economic models 3 Maimization 31 Single variable maimization 1 3 Multi variable maimization 6 33 Concave unctions 9 34 Maimization with non-negativity constraints 9 14 pages

2 31 Single variable maimization Suppose that the unction : takes on its maimum at From the diagram it is clear that i the unction is dierentiable at, then the slope o the graph o ( ) must be zero at Tangent line That is, the ollowing condition is necessary or a maimum First order necessary condition or a maimum Suppose that ( ) takes on its maimum at I is dierentiable at then ( ) While a igure is worth a thousand words, it does not constitute a proo But by using the deinition o a derivative as a limit, this result is quite easy to demonstrate We show that i the slope is not zero, then the unction does not take on its maimum at Proo: Suppose that the slope o the graph o is strictly positive at, ie ( ) This is depicted in the let igure below Since the derivative is the limit o the slope o chords like AB, it ollows that or some suiciently small, ( ) ( ) or all (, ) 1

3 Tangent line Tangent line For (, ) the denominator is strictly positive Thereore ( ) ( ) or all (, ) Thus or all and suiciently close, ( ) ( ) Then does not take on its maimum at By an almost identical argument, it also ollows that i small ˆ such that ( ), there is some suiciently ( ) ( ) or all ˆ ˆ (, ) For (, ) the denominator is strictly negative Thereore ( ) ( ) or all ˆ (, ) Thus or all at and suiciently close, ( ) ( ) Then again does not take on its maimum must be zero We have shown that the slope cannot be strictly positive or strictly negative at so the slope QED

4 By an essentially identical argument, the unction must also have a derivative o zero at the minimizing value o There is a third case as well A unction may have a slope o zero at strictly positive or strictly negative slope on an interval around have a point o inlection at and a In such cases the unction is said to Tangent line Eamples: (i) ( ) 3,, (ii) ( ) (1 ) 5, 1 It is helpul to distinguish between a value o where a unction has a strict local maimum and the value o Deinition: Strict local maimum at where a unction has a global maimum The unction ( ) has a strict local maimum 1 at i ( ) ( ) or all in some interval (, ) The unction ( ) depicted below has a strict local maimum at 1 A unction ( ) has a local maimum 1 at (, ) i ( ) ( ) or all in some interval 3

5 Suicient condition or a strict local maimum at I ( ) and ( ) then ( ) has a strict local maimum at We now sketch a proo o this result Note that i strictly decreasing at Since d ( ) ( ) then the derivative ( ) is d ( ) it ollows that or some interval (, ) the derivative ( ) is strictly negative Thereore ( ) is strictly decreasing or all in (, ) Also, since the derivative ( ) is strictly decreasing and is zero at, it ollows that or some interval the derivative is strictly positive Thereore ( ) is strictly increasing or all in (, ) (, ) QED By an almost identical argument we have the ollowing urther result Suicient conditions or a strict local minimum at I ( ) and ( ) then ( ) has a strict local maimum at 4

6 I ( ) takes on its global maimum at, then ( ) cannot have a local minimum at From the suicient conditions or a local minimum the second derivative cannot be strictly positive We thereore have the ollowing urther necessary condition or a maimum Second order (necessary) condition or a maimum Suppose that the unction : takes on its maimum at I the unction is twice dierentiable at then d ( ) d Maimization o a concave unction I a unction is concave and dierentiable it is especially easy to solve or the maimizing value For the one variable case we argued that i a unction is dierentiable, then a concave unction can be deined as ollows: Deinition : Dierentiable Concave unction : The dierentiable unction ( ) is concave on S i or all and S ( ) ( ) ( )( ) Remember that y ( ) ( )( ) is a line with slope ( ) through the point (, ( )) A unction is concave i every such tangent line lies above the graph o the unction Suppose that ( ) It ollows immediately rom Deinition, that or all and S ( ) ( ) ( )( ) ( ) 5

7 Thus i the First Order Condition holds at and the unction is concave, then ( ) takes on its (global) maimum at Tangent line Concave unction 3 Multi variable maimization n Suppose that the unction : takes on its maimum at Then varying only the -th component o the vector, (,,,,,, ) must take on its maimum at n Appealing to the proo in the one variable case, we then have the ollowing two sets o necessary conditions or a maimum First order necessary conditions or a maimum Suppose that ( ) takes on its maimum at I is dierentiable at then ( ), 1,, n 6

8 Tangent line Necessary conditions or a maimum Second order (necessary) conditions or a maimum n Suppose that the unction : takes on its maimum at dierentiable at then I the unction is twice ( ), 1,, n Deriving suicient conditions is more complicated To understand the issues consider the ollowing unction: ( ) As you can readily check, the gradient vector is ( ) ( 1 6,61 ) Thus the First Order Conditions are satisied at (,) Note net that (,) 4 and 1 1 (, ) 4 Thus along each ais the unction has a maimum at the origin 7

9 The unction ( ) is depicted below However suppose we choose 1 t Then ( ( t), ( t)) 4 ( t) ( t) 6 ( t) ( t) 4 4t Thus the unction does not have a maimum at (,) 33 Concave unctions As in the one variable case, solving or a maimum is easier i the unction to be maimized is concave We appeal to the ollowing deinition o a dierentiable concave unction 8

10 n Deinition : Dierentiable Concave unction : The dierentiable unction ( ) is concave on S i or all and S ( ) ( ) ( ) ( ) It ollows immediately that i the gradient vector is zero at, then ( ) ( ) and so ( ) takes on its maimum at 3 Maimization with non-negativity constraints Typically economic models deal with variables that are non-negative Then the maimization problem is Ma{ ( ) } Suppose that the solution o this maimization problem is must solve the ollowing problem Fiing all but the -th component o, Ma{ (,,,,,, ) } n Case (i) This is depicted below Arguing as in the previous section, the -th partial derivative must be zero at 9

11 Tangent line Necessary condition or a maimum Case (ii) I, as depicted below, ( ), then or all in some interval (, ) (,,,,,, ) (,,,,,, ) n n Tangent line not maimized at 1

12 Thus ( ) does not take on its maimum at Then the necessary condition or a maimum when is that ( ) Combining the two cases yields the ollowing necessary conditions or a maimum First Order Conditions with non-negativity constraints Suppose that solves Ma{ ( ) } Then ( ), 1,, n Moreover i the inequality is strict then Note that the solution must satisy the ollowing two inequality constraints: ( ) and, 1,, n However i one o these constraints is slack ie not binding, then the other must be binding, that is ( ), 1,, n For this reason the necessary conditions are oten reerred to as the complementary slackness conditions Practical tip I you are solving a maimization problem with non-negativity constraints, unless you have reason to believe a particular variable may be zero, begin by assuming that conditions are the ollowing system o equality constraints: Then the necessary ( ), 1,, n 11

13 Suppose you solve or satisying these conditions and ind that, or some i, easible But it suggests that the non-negativity constraint will be binding or the proceed under this assumption by setting variables i i i-th and solving the system o FOC or the other This is not variable Then n 1 I the unction ( ) is concave, then the necessary conditions are also suicient or a maimum Otherwise you will still have to check whether the necessary conditions hold when one or more components o are zero Eample: Multi-product irm The total cost o producing output vector q ( q1, q) is C( q, q ) q q 6q q A irm is a price-taker in both o its output markets The price vector is p (8,8) The total proit is ( q) 8q 8q 6q q We irst seek a solution under the assumption that * q I so the FOC are as ollows: ( q) 8 q1 6q 1 ( q) 8 q 6q1 As is easily conirmed, the solution o these two equation is q (1,1) and the resulting proit is ( q ) 8 Note that 1 Thus, holding q constant, the slope o the graph o the proit unction declines as q 1 increases By the same argument, holding q 1 constant, the slope o the graph o the proit unction declines as q increases It is thereore tempting to think that we have ound the solution 1

14 However we need to check whether there is also a solution with conditions are as ollows: 1 In this case the necessary ( q) 8 q1 6q 1 ( q) 8 q 6q1, where we have the equality since q Setting q1 the FOC can be rewritten as ollows: ( q) 86q 1 ( q) 8 q, From the second o these conditions, q * 4 Then 1 * ( q ) so the irst o these conditions is also satisied Thus * q (, 4) also satisies the necessary conditions or a maimum Note that * ( q ) 16 ( q ) Thus, the irm is strictly better o producing only product 1 than both products The proit unction is depicted below The point q (1,1) at which the First Order Conditions hold is not even a local maimum It is called a saddle point ** Given the symmetry o the problem, q (4,) must also satisy the necessary conditions 13

15 (1,1) 14