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
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
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
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
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
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
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 1 1 1 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
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: ( ) 4 6 1 1 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
The unction ( ) is depicted below However suppose we choose 1 t Then ( ( t), ( t)) 4 ( t) ( t) 6 ( t) ( t) 4 4t 1 1 1 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
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{ (,,,,,, ) } 1 1 1 n Case (i) This is depicted below Arguing as in the previous section, the -th partial derivative must be zero at 9
Tangent line Necessary condition or a maimum Case (ii) I, as depicted below, ( ), then or all in some interval (, ) (,,,,,, ) (,,,,,, ) 1 1 1 n 1 1 1 n Tangent line not maimized at 1
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
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 1 1 1 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 1 1 1 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
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
(1,1) 14