ECON0702: Mathematical Methods in Economicsyluo/teaching/lecture4.pdfnow is: How can we make the...

ECON0702: Mathematical Methods in Economics

Yulei Luo

SEF of HKU

February 15, 2009

Luo, Y. (SEF of HKU) MME February 15, 2009 1 / 81

Constrained Static Optimization

So far we have focused on �nding the maximum or minimum value ofa function without restricting the choice variables. In many economicproblems, the choice variables must be constrained by economicconsiderations.

E.g., consumers maximizes their utility functions with the budgetconstraints. A �rm minimizes the cost of production with theconstraint of production technology.

Such constraints may lower the maximum (or increase the minimum)for the objective function being maximized (or minimized). Becausewe are not able to choose freely among all choice variables, theobjective function may not be as large as it can be.

The constraints would be said to be nonbinding if we could obtain thesame level of the objective function with or without imposing theconstraint.


Finding the Stationary Values

For illustration, consider a consumer�s choice problem: maximize hisutility as follows

u (x1, x2) = x1x2 + 2x1, (1)

subject to the budget constraint

4x1 + 2x2 = 60. (2)

For this simple constrained problem, we can solve it by substitutingthe budget constraint into the objective utility function without usingany new technique: The budget constraint can be rewritten asx2 = 60�4x1

2 = 30� 2x1,which combining with the utility functiongives

u (x1) = x1 (30� 2x1) + 2x1. (3)

By setting ∂u∂x1= 32� 4x1 = 0, we get x�1 = 8 and x�2 = 14. Since

d 2udx 21

= �4 < 0, the stationary value constitutes a constrainedmaximum.Luo, Y. (SEF of HKU) MME February 15, 2009 3 / 81

The Lagrange multiplier method

However, when the constraint is itself a complicated function or whenthe constraint cannot be solved to express one variable as an explicitfunction of the other variables, the technique of substitution andelimination is not enough to solve the constrained optimizationproblem. We will introduce a new method: the Lagrange multipliermethod to solve the constrained optimization problem.The essence of the LM method is to convert a constrained extremumproblem into a form such that the FOC of the free extremum problemcan still be applied.In general, given an objective function

z = f (x , y) (4)

subject to the constraint

g (x , y) = c where c is a constant, (5)

we can write the Lagrange function as follows

Z = L (x , y ,λ) = f (x , y) + λ [c � g (x , y)] . (6)Luo, Y. (SEF of HKU) MME February 15, 2009 4 / 81

(conti.) The symbol, λ, is called a Lagrange multiplier and will bedetermined and discussed later.If c � g (x , y) = 0 always holds, the last term of Z will vanishregardless of the value of λ. Hence, �nding the constrained maximumvalue of z is equivalent to �nding a critical value of Z . The questionnow is: How can we make the parenthetical expression in (6) vanish?Let�s proceed to do so, treating λ also as an additional choice variable(in addition to x and y). From the Lagrange function (6), the FOCsare

∂L∂x

= fx � λgx = 0, (7)

∂L∂y

= fy � λgy = 0, (8)

∂L∂λ

= c � g (x , y) = 0. (9)

The �nal equation will automatically guarantee the satisfaction of theconstraint. Since λ [c � g (x , y)] = 0, the stationary values of Z in(6) must be identical with those of (4), subject to (5).


Reconsider the above consumer�s choice problem, �rst, we can writethe Lagrange function as follows

Z = L (x1, x2,λ) = x1x2 + 2x1 + λ [60� (4x1 + 2x2)] . (10)

The FOCs are:

∂L∂x1

= x2 + 2� 4λ = 0, (11)

∂L∂x2

= x1 � 2λ = 0, (12)

∂L∂λ

= 60� (4x1 + 2x2) = 0, (13)

solving the above equation for the critical values gives

x�1 = 8, x�2 = 14, and λ = 4. (14)


Summary of the Procedure

Step 1: Forming the Lagrange function

Z = L(x , y ,λ) = f (x , y) + λ[c � g(x , y)]. (15)

Step 2: Find the critical points of the Lagrangian function L(x , y ,λ)by computing ∂L/∂x , ∂L/∂y , and ∂L/∂λ and setting each equal to 0to solve for optimal (x�, y �,λ�) :

∂L∂x= 0,

∂Ly= 0, and

∂L∂λ= 0.

Note that since λ just multiplies the constraint in the de�nition of L,the equation ∂L/∂λ = 0 is equivalent to the constraintc � g(x , y) = 0.Note that by introducing the Lagrange multiplier λ into theconstrained problem, we have transformed a two-variable constrainedproblem to the three-variable unconstrained problem of �nding thecritical points of a function L(x , y ,λ).


Total Di¤erential Approach

In the discussion of the free extremum of z = f (x , y), it is learnedthat the necessary FOC can be stated in terms of the total di¤erentialdz :

dz = fxdx + fydy = 0. (16)

This statement remains valid after adding the constraint g(x , y) = c .

However, with the constraint, we can no longer take both dx and dyas arbitrary change as before because now dx and dy are dependenteach other:

g(x , y) = c =) (dg =) gxdx + gydy = 0. (17)


(conti.) The FOC in terms of total di¤erential becomes

dz = 0 subject to g(x , y) = c and gxdx + gydy = 0. (18)

In order to satisfy this necessary FOC, we must have

fxgx=fygy, (19)

which together with the constraint g(x , y) = c will provide twoequations to solve for the critical values of x and y .

Hence, the total di¤erential approach yields the same FOC conditionsas the Lagrange multiplier method. Note that the LM method givesthe value of λ� as a direct by-product.


An Interpretation of the Lagrange Multiplier

The Lagrange multiplier λ measures the sensitivity of Z � (which isthe value of Z in optimum) to the change in the constraint. In otherwords, it gives us a new measure of value of the scarce resources (Thee¤ect of an increase in c would indicate how the optimal solution isa¤ected by a relaxation of the constraint). If we can express theoptimal values of (x�, y �,λ�) as implicit functions of c :

x� = x� (c) ; y � = y � (c) ;λ� = λ� (c)

and all of which have continuous derivatives. Further, we have thefollowing identities:

fx (x�, y �)� λ�gx (x�, y �) = 0, (20)

fy (x�, y �)� λ�gy (x�, y �) = 0, (21)

c � g (x�, y �) = 0. (22)


(conti.) and the Lagrange function in optimum can be written as

Z � = L(x�, y �,λ�) = f (x�, y �) + λ�[c � g(x�, y �)], (23)

which means that

dZ �

dc= fx

dx�

dc+ fy

dy �

dc+dλ�

dc[c � g(x�, y �)| {z }

=0

]

+λ��1�

�gxdx�

dc+ gy

dy �

dc

��

=

0B@ fx � λ�gx| {z }=0 (FOC for x )

1CA dx�

dc+

0B@ fy � λ�gy| {z }=0 (FOC for x )

1CA dy �

dc+ λ�

= λ�.


Generalization: n-Variable and Multi-constraint Case

The optimization problem can be formed as follows:

maxfx1,��,xng

�min

fx1,��,xng

�z = f (x1, � � �, xn), (24)

subject to : g(x1, � � �, xn) = c (25)

It follows that the Lagrange function is

Z = L (x1, � � �, xn,λ) = f (x1, � � �, xn) + λ [c � g(x1, � � �, xn)] , (26)

for which the FOCs are

fi (x1, � � �, xn) = λgi (x1, � � �, xn), i = 1, � � �, n.g(x1, � � �, xn) = c


(conti.) If the objective function has more than one variable, say, twoconstraints

g(x1, � � �, xn) = c and h(x1, � � �, xn) = d (27)

The Lagrange function is

Z = f (x1, � � �, xn) + λ [c � g(x1, � � �, xn)] (28)

+µ [d � h(x1, � � �, xn)] ,

for which the FOCs are

fi (x1, � � �, xn) = λgi (x1, � � �, xn) + µhi (x1, � � �, xn), i = 1, � � �, n.g(x1, � � �, xn) = c

h(x1, � � �, xn) = d .


Second-Order Conditions

Note that even though Z � is indeed a standard type of extremumw.r.t. the choice variables, it is not so w.r.t. the Lagrange multiplier.(23) shows that unlike (x�, y �), if λ� is replaced by any other value ofλ, no e¤ect will be produced on Z � since c � g(x�, y �) = 0. Thusthe role played by λ� in the optimal solution is di¤ers basically fromthat of x� and y �.

While it is safe to treat λ as another choice variable in the discussionof FOCs, we should treat λ� di¤erently in the discussion of SOCs.

The new SOCs can again be stated in terms of the SO totaldi¤erential d2z . The presence of the constraint will entail certainsigni�cant modi�cations.


Second-Order Total Di¤erential

The constraint g(x , y) = c implies that dg = gxdx + gydy = 0. dxand dy are no longer both arbitrary: we may take dx as an arbitrarychange, but then dy is dependent on dx , i.e., dy = � (gx/gy ) dx .Note that since gx and gy depend on x and y , dy also depends on xand y .Thus,

d2z = d (dz) =∂ (dz)

∂xdx +

∂ (dz)∂y

dy

=

�∂ (fxdx + fydy)

∂x

�dx +

�∂ (fxdx + fydy)

∂y

�dy

=

�fxxdx +

�fxydy + fy

∂dy∂x

��dx +

�fyxdx +

�fyydy + fy

∂dy∂y

��dy

= fxxdx2 + 2fxydxdy + fyydy2 +�fy

∂dy∂xdx + fy

∂dy∂ydy�

= fxxdx2 + 2fxydxdy + fyydy2 + fyd2y


(conti.) The last term disquali�es d2z as a quadratic form. But d2zcan be transformed into a quadratic form by virtue of the constraintg(x , y) = c :

dg = 0 =)d (dg) = gxxdx2 + 2gxydxdy + gyydy2 + gyd2y = 0. (29)

Solving the last equation for d2y and substituting the result in theexpression of d2z gives

d2z =

�fxx �

fygygxx

�dx2 + 2

�fxy �

fygygxy

�dxdy +

�fyy �

fygygyy

�dy2(30)

= (fxx � λgxx ) dx2 + 2 (fxy � λgxy ) dxdy + (fyy � λgyy ) dy2 (31)

= Zxxdx2 + 2Zxydxdy + Zyydy2 (32)

where Zxx = fxx � λgxx ,Zxy = fxy � λgxy , and Zyy = fyy � λgyy arefrom partially di¤erentiating the derivatives in (7) and (8).


Second-Order Conditions

For a constrained extremum problem, the SO necessary and su¢ cientconditions are still determined by the SO total di¤erential d2z for dx anddy satisfying dg = gxdx + gydy = 0.

Theorem (SO su¢ cient conditions)

For maximum of z : d2z negative de�nite, subject to dg = 0. Forminimum of z : d2z positive de�nite, subject to dg = 0.

Theorem (SO necessary conditions)

For maximum of z : d2z negative semide�nite, subject to dg = 0. Forminimum of z : d2z positive semide�nite, subject to dg = 0.


The Bordered Hessian

As in the case of free extremum, it is possible to express the SOsu¢ cient condition in determinantal form. In theconstrained-extremum case we will use what is known a borderedHessian.

Let�s �rst analyze the conditions for the sign de�niteness of atwo-variable quadratic form, subject to a linear constraint:

q = au2 + 2huv + bv2 subject to αu + βv = 0 (33)

Since the constraint means that v = � (α/β) u,

q =�aβ2 � 2hαβ+ bα2

� u2β2, (34)

which means that q is positive (negative) de�nite i¤aβ2 � 2hαβ+ bα2 > 0(< 0).


(conti.) It so happens that the following symmetric determinant��0 α βα a hβ h b

�� = � �aβ2 � 2hαβ+ bα2�. (35)

Consequently, we can state that

q is�positive de�nitenegative de�nite

subject to αu + βv = 0

i¤

��0 α βα a hβ h b

�� =�< 0> 0

. (36)


(conti.) Note that the determinant used in this criterion is nothing

but the discriminant of the original quadratic form

�� a hh b

�� , with aborder placed on top and a similar border on the left: The border ismerely composed of two coe¢ cients α and β from the constraint, plusa zero in the principal diagonal.

When applied to the quadratic form d2z , the (plain) discriminant

consists of the Hessian

�� Zxx ZxyZxy Zyy

�� . Given the constraintgxdx + gydy = 0,

d2z is�positive de�nitenegative de�nite

s.t. dg = 0 i¤

��0 gx gygx Zxx Zxygy Zxy Zyy

��| {z }jH j

=

�< 0> 0

.


Example: Consumer�s Utility Maximization Problem

max u (x1, x2) = x1x2 subject to x1 +x21+ r

= B. (37)

Step1: The Lagrange function is

Z = L (x1, x2,λ) = u (x1, x2) + λ

�B � x1 �

x21+ r

�. (38)

Step 2: The FOCs are

∂Z∂λ

= B � x1 �x21+ r

= 0, (39)

∂Z∂x1

= x2 � λ = 0, (40)

∂Z∂x2

= x1 �λ

1+ r= 0, (41)

which can be used to solve for optimal level of x�1 = B/2 andx�2 = B (1+ r) /2.Luo, Y. (SEF of HKU) MME February 15, 2009 21 / 81

(conti.) Next, we should check the SO su¢ cient condition for amaximum. The bordered Hessian for this problem is

��H�� =��

0 �1 � 11+r

�1 0 1� 11+r 1 0

�� = 21+ r

> 0. (42)

Thus the SO su¢ cient condition is satis�ed for a maximum.


-Variable Case

When the optimization problem takes the form:

maxfx1,��,xng

�min

fx1,��,xng

�z = f (x1, � � �, xn), subject to: g(x1, � � �, xn) = c ,

(43)the Lagrange function is

Z = f (x1, � � �, xn) + λ [c � g(x1, � � �, xn)] , (44)

and dx1, � � �, dxn satisfy the relation:dg = g1dx1 + � � �+ gndxn = 0. (45)

which implies that the bordered Hessian is

��H�� =��0 g1 � � � gng1 Z11 � � � Z1n

� � � � � � . . . � � �gn Zn1 � � � Znn

�� .Luo, Y. (SEF of HKU) MME February 15, 2009 23 / 81

(conti.) its bordered leading principal minors are

��H2�� =

��0 g1 g2g1 Z11 Z12g2 Z21 Z22

�� , ��H3�� =��0 g1 g2 g3g1 Z11 Z12 Z13g2 Z21 Z22 Z23g3 Z31 Z32 Z33

�� ,� � �,

��H�� =��Hn��

Theorem (Conditions for Maximum)

(1) FO necessary condition: Zλ = Z1 = � � � = Zn = 0. (2) SO su¢ cientcondition:

��H2�� > 0, ��H3�� < 0. � ��, (�1)n ��Hn�� > 0.Theorem (Conditions for Minimum)

(1) FO necessary condition: Zλ = Z1 = � � � = Zn = 0. (2) SO su¢ cientcondition:

��H2�� < 0, ��H3�� < 0. � ��, ��Hn�� < 0.Luo, Y. (SEF of HKU) MME February 15, 2009 24 / 81

Quasiconcavity and Quasiconvexity

For an unconstrained optimization problem (a free extremumproblem), the concavity (convexity) of the objective functionguarantees the existence of absolute maximum (absolute minimum).

For a constrained optimization problem, we will show that thequasiconcavity (quasiconvexity) of the objective function guaranteesthe existence of absolute maximum (absolute minimum).

Quasiconcavity (quasiconvexity), like concavity (convexity), can beeither strict or nonstrict.


De�nitionA function is quasiconcave (quasiconvex) i¤ for any pair of distinct pointsu and v in the convex domain of f , and for 0 < θ < 1, f (v) � f (u)implies that

f (θu + (1� θ) v) � f (u) (f (θu + (1� θ) v) � f (v)) . (46)

Further, if the weak inequality ��(��) is replaced by the strictinequality �<�(�>�), f is said to be strictly quasiconcave (strictlyquasiconvex).

FactQuasiconcavity (quasiconvexity) is a weaker condition than concavity(convexity).


Theorem (Negative of a function)

If f (x) is quasiconcave (strictly quasiconcave), then �f (x) is quasiconvex(strictly quasiconvex).

Proof.Use the fact that multiplying an inequality by �1 reverses the sense ofinequality.

Theorem (Concavity vs. quasiconcavity)

Any (strictly) concave (convex) function is (strictly) quasiconcave(quasiconvex) function, but the converse is not true.

Proof.Using the de�nitions of concavity and quasiconcavity to prove.

f (θu + (1� θ) v) � θf (u) + (1� θ) f (v)

� f (u) (because we assume that f (v) � f (u))


Theorem (Linear function)

If f (x) is linear, then it is quasiconcave as well as quasiconvex.

Proof.Using the fact that if f (x) is linear, then it is concave as well asconvex.

FactUnlike concave (convex) functions, a sum of two quasiconcave(quasiconvex) functions is not necessarily quasiconcave (quasiconvex).


Sometimes it is easier to check quasiconcavity and quasiconvexity by thefollowing alternative de�nitions. We �rst introduce the concept of convexsets.

De�nitionIf, for any two points in set S , the line segment connecting these twopoints lies entirely in S , then S is said to be a convex set. A correspondingalgebraic de�nition is: A set S is convex i¤ for any two points u 2 S andv 2 S , and for every scalar θ 2 [0, 1], it is true thatw = θu + (1� θ) v 2 S . Note that u and v can be in the space with anydimension.


FactTo qualify as a convex set, the set of points must contain no holes, and itsboundary must not be indented anywhere. See Figure 11.8 in the book.

FactNote that convex sets and convex functions are distinct concepts. Indescribing a function, the word convex speci�es how a curve or surfacebends itself-it must form a valley. But in describing a set, the wordspeci�es how the points in the set are packed together - they must notallow any holes and the boundary must not be indented.


De�nitionA function f (x), where x is a vector of variables, is quasiconcave(quasiconvex) i¤ for any constant k, the set S� = fx jf (x) � kg(S� = fx jf (x) � kg) is convex.

See Figure 12.5 (a), (b), and (c).The three functions in Figure 12.5 contain concave as well as convexsegments and hence are neither concave or convex.But the function in Fig 12.5 (a) is quasiconcave because for any valueof k, the set S� is convex.The function in Fig 12.5 (b) is quasiconvex because for any value of k,the set S� is convex.The monotonic function in Fig 12.5 (b) is quasiconcave as well asquasiconvex because both S� and S� are convex.

Hence, given that S� is convex, we can only conclude that thefunction f is quasiconcave, but not necessarily concave.Examples: (1) Z = x2 (x � 0) is quasiconvex as well as quasiconcavesince both S� and S� are convex. (2) Z = (x � a)2 + (y � b)2 isquasiconvex since S� is convex.Luo, Y. (SEF of HKU) MME February 15, 2009 31 / 81

If the function z = f (x1, � � �, xn) is twice continuously di¤erentiable,quasiconcavity and quasiconvexity can be checked by the �rst andsecond partial derivatives of the function.

De�ne a bordered determinant as follows:

��B�� =��0 f1 � � � fnf1 f11 � � � f1n

� � � � � � . . . � � �fn fn1 � � � fnn

�� (47)

Note that the determinant��B�� is di¤erent from the bordered Hessian��H��: Unlike ��H��, the border in ��B�� is composed of the �rst derivatives

of the function f rather than the constraint function g .

Hence, if��B�� satis�es the SO su¢ cient condition for strict

quasiconcavity (will be speci�ed below),��H�� must also satisfy the SO

su¢ cient condition for constrained maximization problem.


(conti.) We can de�ne successive principal minors of B as follows:

��B1�� = �� 0 f1f1 f11

�� , ��B2�� =��0 f1 f2f1 f11 f12f2 f21 f22

�� , � � �, ��Bn�� = ��B�� (48)

A su¢ cient condition for f to be quasiconcave on the nonnegativedomain is that��B1�� < 0, ��B2�� > 0, � � �, (�1)n ��Bn�� > 0. (49)

For quasiconvexity, the corresponding condition is that��B1�� < 0, ��B2�� < 0, � � �, ��Bn�� < 0. (50)

A necessary condition for f to be quasiconcave on the nonnegativedomain is that��B1�� 0, ��B2�� 0, � � �, (�1)n ��Bn�� 0. (51)

For quasiconvexity, the corresponding condition is that��B1�� 0, ��B2�� 0, � � �, ��Bn�� 0. (52)


Example: Consider z = f (x1, x2) = x1x2, since

f1 = x2, f2 = x1, f11 = f22 = 0, f12 = f21 = 1,

the relevant principal minors are

��B1�� = �� 0 x2x2 0

�� = �x22 � 0, ��B2�� =��0 x2 x1x2 0 1x1 1 0

�� = 2x1x2 � 0.(53)

Thus, z = x1x2 is quasiconcave on the nonnegative domain.


Example: Show that z = f (x , y) = xayb (x > 0, y > 0, a, b 2 (0, 1))is quasiconcave. Since

fx = axa�1yb , fy = bxayb�1,

fxx = a (a� 1) xa�2yb , fyy = b (b� 1) xayb�2,fxy = fyx = abxa�1yb�1,

��B1�� =

�� 0 fxfx 0

�� = � �axa�1yb�2 < 0, (54)

��B2�� =

��0 fx fyfx fxx fxyfy fyx fyy

�� = 2ab (a+ b) x3a�2y3b�2 > 0, (55)which means that the function is quasiconcave.


(conti.) In this case, the condition for concavity can be expressed as

fxx fyy � f 2xy =ha (a� 1) xa�2yb

i hb (b� 1) xayb�2

i�habxa�1yb�1

i2(56)

= ab (a� 1) (b� 1) x2a�2y2b�2 � a2b2x2a�2y2b�2

= ab [1� a� b] x2a�2y2b�2 (57)

and this expression is positive (as required for concavity) for

1� a� b > 0 =) a+ b < 1.


FactWhen the function in the constraint is linear, that is,g(x1, � � �, xn) = a1x1 + � � �+ anxn = c , the bordered determinant

��B�� andthe bordered Hessian

��H�� have the following relationship:��B�� = λ2��H�� . (58)

Hence, in the linear constraint case, the two bordered determinants alwayshave the same sign at the stationary point.

Fact (Relative vs. absolute extreme)

If a function is quasiconcave (quasiconvex), by the same reasons forconcave (convex) functions, its relative maximum (relative minimum) is anabsolute maximum (absolute minimum).


Utility Maximization and Consumer Demand

Consider the following two-commodity consumer optimizationproblem:

max u (x , y) (ux > 0, uy > 0) (59)

subject to the budget constraint

Pxx + Py y = B (60)

where Px ,Py , and B are given exogenously.

The Lagrange function is then

Z = u (x , y) + λ (B � Pxx � Py y) . (61)


The FOCs are

Zλ = B � Pxx � Py y = 0 (62)

Zx = ux � λPx = 0 (63)

Zy = uy � λPy = 0. (64)

From the last two equations, we have

uxuy=PxPy, (65)

where uxuy= MRSxy is called the marginal rate of substitution (MRS)

of x for y . Thus, we have the well-known equality: MRSxy = PxPy

which is the necessary condition for the interior optimal solution. Seethe �gure of the indi¤erence curve.


If the bordered Hessian is positive, i.e.,��0 Px PyPx uxx uxyPy uyx uyy

�� = 2PxPyuxy � P2y uxx � P2x uyy > 0,in which all elements are evaluated at the optimum, then thestationary value of u is maximum.


Static Optimization with Inequality Constraints

So far we have considered optimization problem with equalityconstraints. Now we shall consider constraints that may be satis�edas inequailities in the solution.Consider the simple optimization problem with inequality constraints:

maxfx ,yg2S

f (x , y) subject to g (x , y) � c . (66)

We seek the largest value attained by f (x , y) in the admissible orfeasible set S of all pairs (x , y) satisfying g (x , y) � c .Note that problems where one wants to minimize f (x , y) subject tofx , yg 2 S can be handled by instead studying the problem ofmaximizing �f (x , y) subject to fx , yg 2 S .This problem can be solved by using an extended Lagrangianmultiplier method introduced above, and it involves examining thestationary points of f in the interior of the feasible set S and thebehavior of f on the boundary of S . This new method is originallyproposed by two Princeton mathematicians: H. W. Kuhn and A.W.Tucker.Luo, Y. (SEF of HKU) MME February 15, 2009 41 / 81

Theorem (Recipe for solving the optimization problem with inequalityconstraints.)

A. Associate a Lagrange multiplier λ with the constraint g (x , y) � c , andde�ne the Lagrangian function as follows

Z = L(x , y) = f (x , y) + λ[c � g(x , y)]. (67)

B. Equate the partial derivatives of Z w.r.t. x and y to zeros:

fx � λgx = 0, fy � λgy = 0. (68)

C. Introduce the complementary slackness condition

λ � 0 and λ[c � g(x , y)] = 0. (69)

D. Require (x , y) to satisfy the constraint

g (x , y) � c (70)


(conti.) Step C (the complementary slackness condition) is tricky. Itrequires that λ is nonnegative and moreover that λ = 0 ifg (x , y) < c . Thus, if λ > 0, we must have g (x , y) = c .

Note that the Lagrange multiplier λ can be interpreted as a price (itis called the shadow price) associated with increasing the right-handside c of the resource constraint g (x , y) � c by 1 unit. With thisinterpretation, prices are nonnegative, and if the resource constraint isnot binding because g (x , y) < c at the optimum, the priceassociated with increasing by one unit is 0.

It is possible to have both λ = 0 and g (x , y) = c . The twoinequalities are complementary in the sense that at most one can be�slack�, that is, at most one can hold with inequality. Equivalently, atleast one must be an equality.


(conti.) Conditions (68) and (69) are called the Kuhn-Tuckerconditions. Note that they are necessary conditions for the aboveproblem.

Note that with an inequality constraint, one will have∂Z∂λ = c � g(x , y) > 0 at an optimum if the constraint holds withinequality at that point. For this reason, we wouldn�t di¤erentiate theLagrangian w.r.t. λ.

Example: Consider the following problem

maxfx ,yg2S

f (x , y) = x2 + y2 + y � 1 (71)

subject to : g (x , y) = x2 + y2 � 1. (72)


Z = x2 + y2 + y � 1 + λ�1�

�x2 + y2

��.


The Trial-and-Error Approach to Search Optimal Solutions

(conti.) The FOCs are then

Zx = 2x � 2λx = 0 (73)

Zy = 2y + 1� 2λy = 0. (74)

The complementary slackness condition is

λ � 0 and λ�1�

�x2 + y2

��= 0. (75)

we want to �nd all pairs (x , y) that satisfy these conditions for somesuitable value of λ.

Begin by looking at (73). This condition implies that

2x (1� λ) = 0.

There are two possibilities: λ = 1 or x = 0. If λ = 1, then (74)implies that 1 = 0, a contradiction. Hence, x = 0.


(conti.) Since x = 0,

Suppose that x2 + y2 = 1, we have y = �1. We try y = 1 �rst.Substituting y = 1 into (74) implies that λ = 3/2 and (75) is satis�ed.Hence, (x , y) = (0, 1) with λ = 3/2 is a candidate for optimality.Similarly, if y = �1, λ = 1/2 and (75) is also satis�ed. Hence,(x , y) = (0,�1) with λ = 1/2 is also a candidate for optimality.Finally, consider the case where x = 0 and x2 + y2 < 1. In this case,(75) implies that λ = 0 and (74) implies y = �1/2. Hence,(x , y) = (0,�1/2) with λ = 0 is also a candidate for optimality.We then conclude that there are 3 candidates for optimality:

f (0, 1) = 1, f (0,�1) = �1, f (0,�1/2) = �5/4.

Because we want to maximize a continuous function over a closedbounded set, by the Extreme value theorem there is a solution to theproblem. Thus, (x , y) = (0, 1) solved the maximization problem.


Consider the n-variable problem

max f (x1, � � �, xn) s.t.

8<:g1 (x1, � � �, xn) � c1

� � �gm (x1, � � �, xn) � cm

(76)

Theorem (Recipe for Solving the General Problem with n-variable)

A). Write down the Lagrange function

L = f (x1, � � �, xn) +m

∑j=1

λj (cj � gj (x1, � � �, xn)) (77)

where λj is the Lagrange multiplier associated with the j � th constraint.


Theorem(conti.) B) Equate all the FOCs to 0, for each i = 1, � � �, n :

∂f∂xi

�m

∑j=1

λj∂gj∂xi

= 0. (78)

C) Impose the complementary slackness conditions:

λj � 0 ( = 0 if cj � gj (x1, � � �, xn) > 0), j = 1, � � �,m. (79)

D) Require x1, � � �, xn to satisfy the constraints

gj (x1, � � �, xn) � cj , j = 1, � � �,m. (80)


Theorem (Kuhn-Tucker Su¢ cient Conditions)

Consider (66) and suppose that (x�, y �) satis�es conditions (68), (69) and(70). If the Lagrange function is concave, then (x�, y �) solves the problem.

Proof.If (x�, y �) satis�es the conditions in (68), then (x�, y �) is a stationarypoint of the Lagrangian. Because a stationary point of the concaveLagrangian will maximize the function, we have

L(x�, y �) = f (x�, y �) + λ[c � g(x�, y �)] � f (x , y) + λ[c � g(x , y)]

Rearranging the terms gives

f (x�, y �)� f (x , y) � λ[g(x�, y �)� g(x , y)] (81)


Proof.(conti.) If c � g (x�, y �) > 0, then by (69), we have λ = 0, so (81) impliesthat f (x�, y �) � f (x , y). On the other hand, if c � g(x�, y �) = 0, then

λ[g(x�, y �)� g(x , y)] = λ[c � g(x , y)].

Here λ � 0, and c � g(x , y) � 0 for all (x , y) satisfying the constraint.Hence, (x�, y �) solves problem (66).


A Special Case: Nonnegativity Conditions on the Variables

Many economic variables must be nonnegative by their very nature. Itis not di¢ cult to incorporate such constraints in our aboveformulation. For example, x � 0 can be expressed ash (x , y) = �x � 0, and we introduce an additional Lagrangemultiplier to go with it.

Consider the problem

maxfx ,yg2S

f (x , y) subject to g (x , y) � c , x � 0, y � 0. (82)

Note that it can be rewritten as

maxfx ,yg2S

f (x , y) subject to g (x , y) � c ,�x � 0,�y � 0. (83)


(conti.) The Lagrange function is then

Z = L(x , y ,λ) = f (x , y) + λ[c � g(x , y)] + µ1x + µ2y . (84)

The FOCs are

fx � λgx + µ1 = 0 (85)

fy � λgy + µ2 = 0 (86)

λ � 0,λ[c � g(x , y)] = 0 (87)

µ1 � 0, µ1x = 0 (88)

µ2 � 0, µ2y = 0, (89)

which is equivalent to

fx � λgx � 0 ( = 0 if x > 0)

fy � λgy � 0 ( = 0 if y > 0)

λ � 0 ( = 0 if c � g(x , y) > 0)


The same idea can obviously be extended to the n-variable problem

max f (x1, � � �, xn) s.t.

8<:g1 (x1, � � �, xn) � c1

� � �gm (x1, � � �, xn) � cm

, (90)

x1 � 0, � � �, xn � 0.

The necessary FOCs for the solution of (90) are that, for eachi = 1, � � �, n :

∂f∂xi

�m

∑j=1

λj∂gj∂xi

� 0 ( = 0 if xi > 0) (91)

λj � 0 ( = 0 if cj � gj (x , y) > 0), j = 1, � � �,m.(92)


Example: The consumer�s maximization problem is

max u = u (x , y)

subject to

Pxx + Py y � B (93)

cxx + cy y � C (94)

x � 0 (95)

y � 0 (96)


Z = u (x , y) + λ1 [B � (Pxx + Py y)] + λ2 [C � (cxx + cy y)]+µ1x + µ2y

Suppose thatu (x , y) = xy2,B = 100,Px = Py = 1,C = 120, cx = 2, and cy = 1.


(conti.) The Kuhn-Tucker conditions are

Zx = y2 � λ1 � 2λ2 � 0, x � 0, xZx = 0.�equivalently, y2 � λ1 � 2λ2 � 0 (= 0 if x > 0)

�Zy = 2xy � λ1 � λ2 � 0, y � 0, yZy = 0.λ1 � 0,λ1 [100� (x + y)] = 0, 100� (x + y) � 0.λ2 � 0,λ2 [120� (2x + y)] = 0, 120� (2x + y) � 0.

Again, the solution procedure involves a certain amount of trial anderror.

We can �rst choose one of the constraints to be nonbinding and solvefor x and y . Once found, use these values to test if the constraintchosen to be nonbinding is violated.If it is, then redo the procedure choosing another constraint to benonbinding. If violation of the nonbinding constraint occurs again, thenwe can assume both constraints bind and the solution is determinedonly by the constraints.


(conti.) Step 1: Assume that the second constraint is nonbinding inthe solution (that is, 120 > 2x + y), so that λ2 = 0 bycomplementary slackness. But let x , λ1, and y be positive so that:

Zx = y2 � λ1 = 0

Zy = 2xy � λ1 = 0

100 = x + y

Solving for x and y yields a trial solution

x = 3313, y = 66

23.

But substituting this into the second constraint gives

2�3313

�+ 66

23= 133

13> 120.

In other words, this solution violates the second constraint and mustbe rejected.


(conti.) Step 2: Reverse the assumption so that λ1 = 0 and let x , λ2,and y be positive so that:

Zx = y2 � λ2 = 0

Zy = 2xy � λ2 = 0

120 = 2x + y

Solving for x and y yields a trial solution

x = 20, y = 80

which implies that λ2 = 3200. These solutions together with λ1 = 0satisfy all constraints. Thus we accept them as the �nal solution tothe Kuhn-Tucker conditions.


Example: Quasi-linear Preference

Consumer�s problem is to choose two commodities to maximize hisutility function:

u(x , y) = y + a ln (x) (97)

subject to

px + qy � I , (98)

x � 0 (99)

y � 0 (100)

where a is a given positive constant. p and q are both positive prices.

First, we construct the Lagrange function as follows

L = y + a ln (x) + λ [I � px � qy ] + +µ1x + µ2y (101)


(conti.) The Kuhn-Tucker conditions can be written as

ax� λp � 0, x � 0,

� ax� λp

�x = 0 (102)

1� λq � 0, y � 0, (1� λq) y = 0 (103)

I � px � qy � 0,λ � 0, (I � px � qy) λ = 0. (104)

in which we have 23 possibilities of equalities or inequalities becausethere are two nonnegative variables and one inequality constraint.First, note that the budget constraint must be binding (that is,px + qy = I , all available income must be used up) because themarginal utility is positive: ux (x , y) > 0 and uy (x , y) > 0, that is,consuming more can result in higher levels of utility. Formally, if theBC is not binding, then λ = 0, which means that ax � 0 and 1 � 0,both are contradiction.Now we can reduce the number of possibilities to 4 :x > 0, x = 0, y > 0, y = 0.We can also rule out the possibility that x = 0 and y = 0 becausethey don�t satisfy the BC: px + qy = I > 0.


(conti.) If x = 0 and y = I/q > 0, the second line in theKuhn-Tucker conditions means that

λ =1q

and the �rst line in the KT conditions implies that

pq� ∞,

which is a contradiction. Intuitively, the initial one unit increase in xwill result in positively in�nite utility, so zero consumption in x can�tbe optimal.


(conti.) If y = 0 and x = I/p > 0, the �rst line in the KT conditionsimplies that

ax� λp = 0 =) λ =

aI. (105)

Substituting it into the second line of the KT conditions gives

1� aIq � 0 =) I � aq.

If these parameters (I , a, q) satisfy this condition, then x = I/p andy = 0 is a candidate optimal solution.

Finally, if both x and y are positive, the �rst two lines in the KTconditions gives

ax� λp = 0 = 1� λq =) x = aq/p and y = I/q � a. (106)

Hence, if I > aq (y > 0), then it is also a candidate optimal solution.


(conti.) In sum, the optimal solution eventually depends on the valuesof parameters (I , a, q):

x = I/p and y = 0 if I � aq.x = aq/p and y = I/q � a if I > aq.


The Envelop Theorem for Unconstrained Optimization

A maximum-value function is an objective function where the choicevariables have been assigned their optimal values. Thus, this functionindirectly becomes a function of the parameters only through theparameters�e¤ects on the optimal values of the choice variables, andis also referred to as the indirect objective function.

The indirect objective function traces out all the maximum values ofthe objective functions these parameters vary. Hence, the IOF is an�envelop� of the set of optimized objective functions generated byvarying the parameters.

Considermax u = f (x , y , φ)

where x and y are choice variables and φ is a parameter.


The FOC necessary conditions:

fx (x , y , φ) = fy (x , y , φ) = 0. (107)

If the SO conditions are met, these two equations implicitly de�ne thesolutions

x� = x� (φ) and y � = y � (φ) .

Substituting these solutions back into the f gives the IOF (or themaximum value function)

V (φ) = f (x� (φ) , y � (φ) , φ) . (108)

Di¤erentiating V (φ) w.r.t. φ gives

dVdφ

= fx∂x�

∂φ+ fy

∂y �

∂φ+ fφ = fφ.

This result means that at the optimum (fx = fy = 0), as φ varies,with x� and y � allowed to adjust, dVdφ gives the same result as if x

�

and y � are treated as constants (only the direct e¤ect need to beconsidered). This is the essence of the Envelop theorem.


The Envelop Theorem for Constrained Optimization

The problem becomes

max u = f (x , y , φ) s.t. g (x , y , φ) = 0.

The Lagrangian is then

Z = f (x , y , φ)� λg (x , y , φ) .

The FOCs are

Zx = fx � λgx = 0

Zy = fy � λgy = 0

Zλ = �g (x , y , φ) = 0.

The optimal solution is then

x� = x� (φ) , y � = y � (φ) ,λ� = λ� (φ) .


Substituting these solutions back into the f gives the IOF (or themaximum value function)

V (φ) = f (x� (φ) , y � (φ) , φ) . (109)

Di¤erentiating V (φ) w.r.t. φ gives

dVdφ

= fx∂x�

∂φ+ fy

∂y �

∂φ+ fφ. (110)

Further, note that since

g (x� (φ) , y � (φ) , φ) = 0,

we have

gx∂x�

∂φ+ gy

∂y �

∂φ+ gφ = 0. (111)

Multiplying λ on both sides and then combining it with the expressionof dVdφ gives the Envelop theorem for constrained optimization:

dVdφ

= (fx � λgx )∂x�

∂φ+ (fy � λgy )

∂y �

∂φ+�fφ � λgφ

�= Zφ. (112)


Homogeneous Functions

De�nitionA function is said to be homogenous of degree r , if multiplication of eachof its independent variables by a constant j will alter the value of thefunction by the proportion j r , that is,

f (jx1, � � �, jxn) = j r f (x1, � � �, xn) . (113)

In economics applications, we assume that j is usually taken to be positive.

Example

Given the function f (x , y ,w) = x/y + 2w/3x , if we multiply eachvariable by j , we get

f (jx , jy , jw) = x/y + 2w/3x = j0f (x , y ,w) , (114)

which means that this function is homogenous function of degree 0(j0 = 1).


De�nitionProduction functions are usually homogeneous functions of degree 1. Theyare often referred to as linearly homogeneous functions.

Assume that the production function has the following form:

Q = f (K , L) (115)

The mathematical assumption of linear homogeneity would amount tothe economic assumption of constant returns to scale (CRTS),because linear homogeneity means that raising all inputs j-fold willalways raise the output (value of the function) exactly j-fold also.


FactGiven the LH production function Q, the average production of labor(APL) and of capital (APK) can be expressed as functions of thecapital-labor ratio, k = K/L alone.

FactMultiplying each independent variable by a factor j = 1/L and using theproperty of linear homogeneity, we have

APL =QL= f

�KL, 1�= f (k, 1) = φ (k) . (116)

APK =QK=QLLK=

φ (k)k

(117)

Therefore, while the production function is homogeneous degree one, bothAPL and APK are homogenous of degree zero in the K and L.


FactGiven the LH production function Q, the marginal production of labor(MPL) and of capital (MPK) can be expressed as functions of thecapital-labor ratio, k = K/L alone.

FactTo �nd the marginal products, we rewrite the total product asQ = Lφ (k) .Di¤erentiating it w.r.t. K and L gives

MPK =∂Q∂K

= L∂φ (k)

∂K= L

dφ (k)dk

∂k∂K

= φ0 (k)

MPL =∂Q∂L

= φ (k) + L∂φ (k)

∂L= φ (k) + Lφ0 (k)

��KL2

�= φ (k)� φ0 (k) k

They are also homogenous of degree zero in the K and L (they remain thesame as long as k is held constant.


Theorem (Euler�s Theorem)

If Q = f (K , L) is linearly homogeneous, then

K∂Q∂K

+ L∂Q∂L

= Q.

Proof.

K∂Q∂K

+ L∂Q∂L

= Kφ0 (k) + L�φ (k)� φ0 (k) k

�= Lφ (k) = Q.

This theorem says that the value of a LH function can always be expressedas a sum of terms, each of which is one of the independent variables andthe FO partial derivative w.r.t. to that variable. Hence, under conditionsof CRTS, if each input is paid the amount of its marginal product, thetotal product will be exhausted by the distributive shares for all the inputfactors, or the pure economic pro�t will be zero.


Example (Cobb-Douglas Production Function)One speci�c production function is the Cobb-Douglas function:

Q = AK αLβ (118)

where A is a positive constant, α and β are positive fractions. Majorfeatures of this production are:(1) It is homogenous of degree (α+ β) .(2) In the special case of α+ β = 1, it is linearly homogeneous.(3) Its isoquants are negatively sloped throughout and strictly convex forpositive values of K and L.(4) It is strictly quasiconcave for positive K and L.


Example (Cobb-Douglas Production Function)

(conti.) For the special case α+ β = 1,

Q = AK αL1�α = ALkα. (119)

APL =QL= Akα,APK =

QK= Akα�1. (120)

MPK =∂Q∂K

= Aαkα�1,MPL =∂Q∂L

= A (1� α) kα. (121)

The Euler theorem can be veri�ed as follows:

K∂Q∂K

+ L∂Q∂L

= KAαkα�1 + LA (1� α) kα

= kα [AαL+ A (1� α) L] = ALkα = Q.


Example (CES Production Function)

A constant elasticity of substitution (CES) production function is

Q = A�δK�ρ + (1� δ) L�ρ

��1/ρ (A > 0; 0 < δ < 1;�1 < ρ 6= 0)(122)

where δ is the distributive parameter, like α in the CD function, and ρ isthe substitution parameter-which has no counterpart in the CD function.First, the CES production function is homogeneous of degree one

jQ = Ahδ (jK )�ρ + (1� δ) (jL)�ρ

i�1/ρ. (123)

Second, CD function is a special case of the CES function. When ρ ! 0,the CES function approaches the CD function.


Example (CES Production Function)

(conti.)

Proof.Taking log on both side of the CES production function and then takingthe limitation gives

ln�QA

�= � lim

ρ!0ln [δK�ρ + (1� δ) L�ρ]

ρ

= � limρ!0

�δK�ρ lnK � (1� δ) L�ρ ln LδK�ρ + (1� δ) L�ρ

= ln�K δL1�δ

�. =)

Q = AK δL1�δ.

in which we use the L�Hopital�s rule.


Least-Cost Combination of Inputs (Application of theHomogenous PF)

Minimizing cost problem:

minfa,bg

C = aPa + bPb (124)

subject to the output constraint

Q (a, b) = Q0 (125)

where Q0,Pa, and Pb are given exogenously. The marginal productsare positive: Qa > 0,Qb > 0.

The Lagrangian can be written as

Z = aPa + bPb + µ [Q0 �Q (a, b)] (126)


(conti.) The FOCs are

Zµ = Q0 �Q (a, b) = 0 (127)

Za = Pa � µQa = 0 (128)

Zb = Pb � µQb = 0 (129)

The last two equation imply that

PaQa=PbQb

= µ, (130)

which means that at the optimum, the input price �marginal productratio must be the same for each input. This equality can be rewrittenas

PaPb=QaQb

, MRTab , (131)

where MRTab is the marginal rate of technical substitution of a for b.See Figure 12.8 in CW.


(conti.) Su¢ cient SOC can be used to ensure a minimum cost afterthe FOCs are met. That is, the problem needs a negative borderedHessian:��H�� =

��0 Qa QbQa �µQaa �µQabQb �µQba �µQbb

�� = µ�QaaQ2b � 2QabQaQb +QbbQ2a

�< 0.

(132)Note that the curvature of an isoquant is represented by the secondderivative:

d2bda2

=dda

�dbda

�=dda

��QaQb

�= �

�Qaa +Qab dbda

�Qb �

�Qba +Qbb dbda

�Qa

Q2b

= �

hQaa +Qab

��QaQb

�iQb �

hQba +Qbb

��QaQb

�iQa

Q2b

= � 1Q3b

�QaaQ2b � 2QabQaQb +QbbQ2a

�Hence, the satisfaction of the su¢ cient SOC implies that d

2bda2 > 0

because µ > 0 and Qb > 0. That is, the isoquant is strictly convex.Luo, Y. (SEF of HKU) MME February 15, 2009 78 / 81

Now assume that Q = Aaαbβ. The optimum implies that

PaPb=QaQb

=αbβa=) b�

a�=

β

α

PaPb= a constant

See Figure 12.9. The expansion path serves to describe the least-costcombinations required to produce varying levels of Q0 and is the locusof the points of tangency. In the above case, the EP is a straight line.Note that any homogenous production can give rise to a linear EP.

A more general class of functions, known as homothetic functions,can produce linear EPs too.

De�nitionHomotheticity can arise from a composite function in the form

H = h (Q (a, b))�h0 (Q) 6= 0

�(133)

where Q (a, b) is homogenous of degree r .


(conti.) Note that H (a, b) is in general not homogenous in a and b.Nonetheless, the EPs of it are linear:

Slope of H isoquant = �HaHb

= �h0 (Q)Qah0 (Q)Qb

= �QaQb

= Slope of Q isoquant

= constant for any givenba

given the linearity of the EPs of Q.

Check two examples: H = Q2 (H is also a homogenous function) andH = exp (Q) (H is not a homogenous function).


What is the e¤ect of the change in the exogenous input price ratio PaPb

on the optimal ratio b�a� ? We introduce a concept the elasticity of

substitution:

σ =relative change in

�b�a�

�relative change in

�PaPb

� = d�b�a�

�/�b�a�

�d�PaPb

�/�PaPb

� = d�b�a�

�/d�PaPb

�� b�a��

/�PaPb

� ,

(134)the larger the σ, the greater the substitution between the two inputs.If b

�a� is considered as a function of

PaPb, then σ will be the ratio of a

marginal function to an average function.For the CD function, b

�a� =

βαPaPb, which means that

σ = 1. (135)

For the CES function introduced above,

QLQK

=PLPK

=) K �

L�=

�δ

1� δ

�1/(1+ρ) � PLPK

�1/(1+ρ)

=) σ =1

1+ ρ.


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