ECON0702: Mathematical Methods in Economicsyluo/teaching/lecture3.pdf · 0) as a stationary value...

ECON0702: Mathematical Methods in Economics

Yulei Luo

SEF of HKU

January 21, 2009

Luo, Y. (SEF of HKU) MME January 21, 2009 1 / 46

Static Optimization and Extreme Values

In this topic, we will study goal equilibrium, in which the equilibriumstate is de�ned as the outcome of the optimizing agents (consumers,�rms, and governments). Our primary focus will be on the classicaltechniques -di¤erential calculus- for �nding optimal positions.

The essence of Economics is about how consumers and �rms makeoptimal choices in a certain environment such that limited resourcescan be allocated e¢ ciently across agents and over time.

The most common criterion of choice among alternatives ineconomics is the goal of maximizing sth. or minimizing sth. They canbe categorized under general heading of optimization. From a purelymathematical point of view, maximum or minimum are also calledextreme values.


(conti.) In formulating an optimization problem, we need �rst de�nean objective function in which the dependent variable represents theobject of maximization or minimization and in which the set ofindependent variables represents the objects whose values the agentscan choose. Those independent variables are referred to as choicevariables (also called decision, control, or policy variables).

E.g., a �rm may seek to maximize pro�t, that is, to maximize thedi¤erence between total revenue R and total cost C . Given a marketdemand, R and C are both functions of the output level Q:

π (Q) = R (Q)� C (Q) , (1)

in which π is the objective function and Q is the choice variable.

The optimization problem is to choose Q to maximize π:

π� = maxQ

π (Q) = R (Q)� C (Q) , (2)

where π� is the optimal level of π, i.e., π �s maximal level.


Relative Maximum and Minimum: First-Derivative Test

We consider the optimizing problem with the general functiony = f (x),and attempt to develop a procedure to �nd the optimalvalue of x that will maximize or minimize y .

See Figure 9.1. (c) in CW. The points E and F are relative (or local)extremum, in the sense that each of these points represents anextremum in some neighborhood of the point only.

Our discussion will mainly focus on relative extrema since an absolute(or global) extrema must be either a relative extrema or one of theends of the function. If we know all the relative extrema, it is easy toget the absolute extrema by selecting the largest or smallest of theseand comparing it with the end points.


(conti.) For y = f (x), the �rst derivative f 0(x) plays a major role in�nding extreme values. For smooth functions (e.g., f is continuouslydi¤erentiable. For non-smooth functions, the derivative doesn�t existat some point), relative extreme values can be only occur where

f 0(x) = 0, (3)

which is a necessary (but not su¢ cient) condition for a relativeextremum (either maximum or minimum).


Theorem (First-derivative test for relative extremum)

If f 0(x0) = 0, then the value of the function at x0, f (x0), will be (a) Arelative maximum if f 0(x) changes its sign from positive to negative fromthe immediate left of the point x0 to immediate right. (b) A relativeminimum if f 0(x) changes its sign from negative to positive from theimmediate left of x0 to immediate right. (c) No extreme point if f 0(x) hasthe same sign on some neighborhood.

We call the value of x0 a critical value of x if f 0(x0) = 0, and refer tof (x0) as a stationary value of y . The point with coordinates x0 andf (x0) can be called a stationary point (wherever the slope is 0, thepoint in question is never situated on an upward or downward incline,but is rather at a standstill point).


(conti.) Example: For

y = f (x) = (x � 1)3 , (4)

x = 1 is not an extreme point even f 0(1) = 0 since f 0(x) has thesame sign on some neighborhood of 1.

Example: For

y = f (x) = x3 � 12x2 + 36x + 8, (5)

setf 0(x) = 3x2 � 24x + 36 = 0 (6)

to get the critical values. Solving it gives x = 2 or 6.It is easy toverify that f 0(x) > 0 for x < 2, and f 0(x) < 0 for x > 2. Thus x = 2is a maximum point and the corresponding maximum value of thefunction f (2) = 40. Similarly, we can also verify that x = 6 is aminimum point.


Second and Higher Derivatives

The derivative of f 0(x) is called second derivative. Similarly, we can�nd derivatives of even higher orders. These will enable us to developalternative criteria for locating the relative extrema of a function.

The SD of f is denoted by f 00 or d2y/dx2. If f 00 exists for all x , f (x)is said to be twice di¤erentiable; if f 00 is continuous, f (x) is said tobe twice continuously di¤erentiable.

The higher order derivatives can be obtained and symbolized alongthe same line:

f 000 (x) , f (4) (x) , � � �, f (n) (x) or (7)

d3ydx3

,d4ydx4

, � � �, dnydxn

. (8)

Example: y = f (x) = 4x4 � x3 + 17x2 + 3x � 1, thenf 0 (x) = 16x3 � 3x2 + 34x + 3, f 00 (x) = 48x2 � 6x + 34, f 000 (x) =96x � 6, f (4) (x) = 96, f (5) (x) = 0.Luo, Y. (SEF of HKU) MME January 21, 2009 8 / 46

Interpretation of the Second Derivative

The second derivative f 00 measures the rate of change of the �rstderive f 0. In other words, f 00 measures the rate of change of the rateof change of the function f .

With a given in�nitesimal increase in x from x = x0,

f 0(x0) > 0f 0(x0) < 0

�means that the value of f tends to

�increasedecrease

;

f 00(x0) > 0f 00(x0) < 0

�means that the slope of f tends to

�increasedecrease

.

For example, when f 0(x0) > 0 and f 00(x0) > 0, the slope of the curveis positive and increasing. Likewise, f 0(x0) > 0 and f 00(x0) < 0means that the slope of the curve is positive but decreasing �thevalue of the function is increasing at a decreasing rate. See Figure 9.5(a) and (b) in the book.


(conti.) f 00(x0) relates to the curvature of a graph; it determines howthe curve tends to bend itself. There are two di¤erent types ofcurvature:

Strictly concave (Figure 9.5 (a))Strictly convex (Figure 9.5 (b))A function whose graph is strictly concave (strictly convex) is called astrictly concave (strictly convex) function.

Precise geometric characterization of a strictly concave (convex)function is as follows. If we pick any pair of points M and N on itscurve and joint them by a straight line, the line segment MN must lieentirely below (above) the curve, except at points M and N.

Note that if the condition is relaxed somewhat, so that the linesegment MN is allowed to either below (above) the curve or along(coinciding with) the curve, then we can describe instead a concave(convex) function, without the adverb �strictly�.


(conti.) Note that a strictly concave or convex curve can nevercontain a linear segment anywhere.

While a strictly concave (convex) function is automatically a concave(convex) function, the converse is not true.

Example: Consider a quadratic function

y = ax2 + bx + c (a 6= 0). (9)

it is obvious that the second derivative of y w.r.t. x , a, determineswhether this function will have a strictly convex (U-shaped) or astrictly concave (inverse U-shaped) graph.


Application to the Expected Utility Model

Assume that a potential gamble player has a strictly concave utilityfunction

u = u (x)

where x is the payo¤, with u (0) = 0, u0 (x) > 0 (positive marginalutility of income or payo¤), and u00 (x) < 0 (diminishing marginalutility) for all x .The decision facing this player involves the choice between twoactions: First, by not playing the game, he saves $15 and enjoys theutility u (15) . Second, by playing, the person has a 1/2 probability ofreceiving $10 and thus enjoying u(10), plus a 1/2 probability ofreceiving $20 and thus enjoying u(20). The expected utility fromplaying is

EU =12u (10) +

12u (20) , (10)

which is less than u (15) since the utility function is strictly concave(See Figure 9.6 (a)). Hence, this risk averse player will always avoidthis game.Luo, Y. (SEF of HKU) MME January 21, 2009 12 / 46

Theorem (Second-derivative test for relative extremum)

If f 0(x0) = 0, then the value of the function at x0, f (x0), will be (a) Arelative maximum if f 00(x0) < 0. (b) A relative maximum if f 00(x0) > 0.

Note that this test is more convenient to use than the �rst derivative testsince it doesn�t require us to check the derivative sign to both the left andright of x .

Example: Given that y = f (x) = 4x2 � x , f 0 (x) = 8x � 1 andf 00 (x) = 8 > 0, we know f (x) reaches its minimum at x� = 1/8.Example: Given that y = g (x) = x3 � 3x2 + 2, we haveg 0 (x) = 3x2 � 6x and g 00 (x) = 6x � 6. Setting g 0 (x) = 0, weobtain two critical values x = 0 or 2, which in turn yield the twoextreme values g (0) = 2 (a maximum because g 00 (0) = �6 < 0)and g (2) = �2 (a minimum because g 00 (0) = 6 > 0).


Necessary versus Su¢ cient Conditions

As was the case with the �rst derivative test, f 0(x) = 0 also plays therole of a necessary condition in the second derivative test. Since thiscondition is based on the FO derivative, it is often referred to as the�rst order condition (FOC).

Once we �nd the FOC satis�ed at x = x0, the negative (positive) signof f 00(x0) is su¢ cient to establish the stationary value in question as arelative maximum (minimum). These su¢ cient conditions are oftenreferred to as second order conditions (SOC).

Note that a relative maximum (minimum) can occur not only whenf 00(x0) is negative (positive), but also when f 00(x0) is zero.Consequently, second order necessary conditions must be couched interms of weak inequalities: for a stationary value f (x0) to bemaximum (minimum), it is necessary that f 00(x0) � 0(� 0).


Condition for Pro�t Maximization

Suppose that R = R (Q) is the total revenue function andC = C (Q) is the total cost function where Q is the level of output.The pro�t function is de�ned as

π (Q) = R (Q)� C (Q) . (11)

The pro�t-maximizing output Q� must satisfy

π0 (Q�) = R 0 (Q�)� C 0 (Q�) = 0 or R 0 (Q�) = C 0 (Q�) , (12)

which means that the marginal revenue (MR) of output is equal tothe marginal cost (MC ) of output. To be sure that the above FOCleads to a maximum, we require:

π00 (Q�) = R 00 (Q�)� C 00 (Q�) < 0, (13)

which means that if the rate of change of MR is less than the rate ofchange of MC at Q�, then output Q� will maximize pro�t.


(conti.) Example: Given that R (Q) = 1200Q � 2Q2 andC (Q) = Q3 � 61.25Q2 + 1528.5Q + 2000, then the pro�t function is

π (Q) = �Q3 + 59.25Q2 � 328.5Q � 2000 (14)

and setting the π0 (Q) = 0 gives the �rst order condition (FOC)

π0 (Q) = �3Q2 + 118.5Q � 328.5 = 0, (15)

which means thatQ� = 3 or 36.5. (16)

Since

π00 (Q�) = �6Q� + 118.5 =�

100.5 > 0 when Q� = 3�100.5 < 0 when Q� = 36.5 ,

(17)the pro�t-maximizing output is Q� = 36.5.


Taylor Series and the Mean-Value Theorem

Now we consider how to expand a function y = f (x) around anypoint x = x0 into what is known as Taylor series. This expansionmeans to transform the function into a polynomial form, in which thecoe¢ cients of the various terms are expressed in terms of thederivative values f 0 (x0) , f 00 (x0) , etc. evaluated at the point ofexpansion x0.

Theorem (Taylor�s Theorem)

Given an arbitrary function y = f (x) , if we know the values of allderivatives: f 0 (x0) , f 00 (x0) , etc., then this function can be expandedaround the point x0 as follows:

f (x) = f (x0) + f 0 (x0) (x � x0) +f 00 (x0)2!

(x � x0)2 (18)

+ � � �+ f(n) (x0)n!

(x � x0)n + Rn= Pn + Rn (19)


Theorem (conti.)

where n! = n(n� 1) � � � 1 is the n factorial, Pn is the n-th degreepolynomial and Rn denotes a remainder which can be expressed in theLagrange form:

Rn =f (n+1) (p)(n+ 1)!

(x � x0)n+1 (20)

where p is some number between x and x0.

Note that when n = 0, the Taylor series reduce to the so-calledmean-value theorem.


Theorem (Mean-value Theorem)

f (x) = f (x0) + f 0 (p) (x � x0) or f (x)� f (x0) = f 0 (p) (x � x0) , (21)

which means that the di¤erence between the value of the function f at x0and at any other x value can be expressed as the product of the di¤erencex � x0 and f 0 (p) with p being some point between x and x0.

If x0 = 0, the Taylor series reduce to the Maclaurin series:

f (x) = f (0) + f 0 (0) x +f 00 (x0)2!

x2 (22)

+ � � �+ f(n) (x0)n!

xn +f (n+1) (p)(n+ 1)!

xn+1, (23)

where p 2 (0, x) .


Nth-Derivative Test

A relative extremum of the function f can be equivalently de�ned asfollows:

De�nitionA function f (x) attains a relative maximum (minimum) value at x0 iff (x)� f (x0) is nonpositive (nonnegative) for values of x in someneighborhood of x0.

Suppose that f (x) has �nite, continuous derivatives up to the desiredorder at x = x0, then the function can be expanded around x = x0 as aTaylor series:

f (x) = f (x0) + f 0 (x0) (x � x0) +f 00 (x0)2!

(x � x0)2 (24)

+ � � �+ f(n) (x0)n!

(x � x0)n +f (n+1) (p)(n+ 1)!

(x � x0)n+1 ,

from which we have the Nth-Derivative Test.Luo, Y. (SEF of HKU) MME January 21, 2009 20 / 46

Theorem (Nth-Derivative Test)

If f 0 (x0) = 0, and if the �rst nonzero derivative value at x0 encountered insuccessive derivation is that of the Nth derivative, f (N ) (x0) 6= 0, then thestationary value f (x0) will be (a) A relative maximum if N is an evennumber and f (N ) (x0) < 0. (b) A relative minimum if N is an even numberand f (N ) (x0) > 0. (c) An in�ection point if N is odd.

Example: Given that y = f (x) = (7� x)4 , since

f 0 (7) = �4 (7� 7)3 = 0; (25)

f 00 (7) = 12 (7� 7)2 = 0; (26)

f 000 (7) = �24 (7� 7) = 0; (27)

f (4) (7) = 24 > 0, (28)

x� = 7 is a minimum point such that f (7) = 0.


Derivatives of Exponential and Logarithmic Functions

Both functions have important applications in economics, especially ingrowth theory and economic dynamics.

The exponential function may be represented in the form:

y = f (t) = bt (b > 1) (29)

where b is a �xed base of the exponent. Its generalized form:y = abct .

If the base is the irrational number denoted by e = 2.71828..., thefunction

y = aert

is referred to the natural exponential function, which can bealternatively written as

y = a exp (rt) (30)


(conti.) For the exponential function and the natural exponentialfunction, taking the log of to the base and the base, respectively, wehave

t = logb y and t = loge y = ln y . (31)

Some rules:

ln (uv) = ln u + ln v (log of product) (32)

ln (u/v) = ln u � ln v (log of quotient) (33)

ln (ua) = a ln u (log of power) (34)

logb u = (logb e) (loge u) (conversion of log base) (35)

logb e = 1/ loge b (inversion of log base) (36)


(conti.) Derivatives of Exponential and Logarithmic Functions

d exp (t)dt

= exp (t) ;d ln (t)dt

=1t

(37)

d exp (f (t))dt

= f 0 (t) exp (f (t)) ;d ln (f (t))

dt=f 0 (t)f (t)

. (38)

The case of Base b :

dbt

dt= bt ln b;

d logb (t)dt

=1

t ln b(39)

dbf (t)

dt= f 0 (t) bf (t) ln b;

d logb (f (t))dt

=f 0 (t)f (t)

1ln b

. (40)


An application: Find dydx from y = xaekx�c . Taking the natural log of

both sides givesln y = a ln x + kx � c . (41)

Di¤erentiating both sides w.r.t. x yields

1ydydx=ax+ k =) dy

dx=� ax+ k

�xaekx�c . (42)


The Case of More than One Choice Variable

The objective is to �nd the relative extreme values of an objectivefunction that involves two or more choice variables.We can express the derivative version of �rst and second condition interms of di¤erentials. Consider the function, we have

dz = f 0 (x) dx . (43)

Since f 0 (x) = 0 is the necessary condition for extreme values, dz = 0is also the necessary condition for extreme values. This FOC requiresthat dz = 0 as x varies.What is the su¢ cient conditions in terms of second-orderdi¤erentials? Di¤erentiating dz = f 0 (x) dx gives

d2z = d (dz) = d�f 0 (x) dx

�= f 00 (x) dx2, (44)

where d2z means that the second di¤erential of z and dx2 means thesquaring of the �rst order di¤erential dx . Hence, d2z < 0(> 0) i¤f 00 (x) < 0(> 0), which means that the second-order su¢ cientcondition for maximum (minimum) of z is d2z < 0(> 0).Luo, Y. (SEF of HKU) MME January 21, 2009 26 / 46

(conti.) For the function z = f (x , y) , the FO necessary condition foran extremum again involves dz = 0 for arbitrary values of dx and dy .In this case, the total di¤erential is

dz = fxdx + fydy . (45)

Thus the equivalent derivative version of the FO necessary conditiondz = 0 is

fx = fy = 0. (46)

To develop a su¢ cient condition, we must look at the SO totaldi¤erential, which is related to the SO partial derivatives.


Second-Order Partial Derivatives

Since the �rst-order derivatives, fx and fy , of z = f (x , y) , arethemselves functions of x and y , we can �nd SO partial derivatives:

fxx =∂

∂xfx or

∂2z∂x2

=∂

∂x

�∂z∂x

�(47)

fyy =∂

∂yfy (48)

fxy =∂2z

∂x∂y=

∂

∂x

�∂z∂y

�(49)

fyx =∂2z

∂y∂x=

∂

∂y

�∂z∂x

�. (50)

where fxy and fyx are called cross (or mixed) partial derivatives. Theyare identical as long as they are both continuous (Young�s theorem).


Second-Order Total Di¤erentials

From the FO total di¤erential,

dz = fxdx + fydy , (51)

we can obtain the SO total di¤erential as follows:

d2z = d (dz) =∂ (dz)

∂xdx +

∂ (dz)∂y

dy

=

�∂ (fxdx + fydy)

∂x

�dx +

�∂ (fxdx + fydy)

∂y

�dy

= (fxxdx + fxydy) dx + (fyxdx + fyydy) dy

= fxxdx2 + 2fxydxdy + fyydy2 (if fxy = fyx ) (52)

Example: Given z = x3 + 5xy � y2,

fxx = 6x , fxy = 5, and fyy = �2 implies thatd2z = 6xdx2 + 10dxdy � 2dy2. (53)


Second-Order Su¢ cient Condition

Using the concept of d2z , we can state the SO su¢ cient condition for:

A maximum of z = f (x , y) if d2z < 0 for any values of dx and dy ,not both zero;

A minimum of z = f (x , y) if d2z > 0 for any values of dx and dy ,not both zero.

The above conditions are only su¢ cient because it is again possible for totake d2z a zero value at a maximum or a minimum. For this reason, SOnecessary conditions must be stated with weak inequalities:

For maximum of z , if d2z � 0 for any values of dx and dy , not bothzero;

For minimum of z , if d2z � 0 for any values of dx and dy , not bothzero.


(conti.) The SO di¤erential conditions can be translated into equivalentcondition on SO derivatives. The following is the main result (A detailedproof requires a knowledge of quadratic form and will be given later.):

d2z < 0 i¤ fxx < 0, fyy < 0, and fxx fyy > (fxy )2

d2z > 0 i¤ fxx > 0, fyy > 0, and fxx fyy > (fxy )2

TheoremConditions for maximum: fx = fy = 0 (FO necessary condition) andfxx < 0, fyy < 0, and fxx fyy > (fxy )

2 (SO su¢ cient condition).

TheoremConditions for minimum: fx = fy = 0 (FO necessary condition) andfxx > 0, fyy > 0, and fxx fyy > (fxy )

2 (SO su¢ cient condition).


(conti.) Example: Find the extreme values ofz = 8x3 + 2xy � 3x2 + y2 + 1. First, we derive all partial derivatives:

fx = 24x2 + 2y � 6x , fy = 2x + 2yfxx = 48x � 6, fyy = 2, fxy = 2.

Second, setting fx = fy = 0 gives

24x2 + 2y � 6x = 0, 2x + 2y = 0, (54)

which gives two solutions for x : x� = 0 or 1/3, and x� = 0 or �1/3.Since fxx (0, 0) = �6 and fyy (0, 0) = 2, it is impossible thatfxx fyy > (fxy )

2 = 4, so that (x�, y �) = (0, 0) is not extreme point.For the solution (x�, y �) = (1/3,�1/3) , we havefxx (1/3,�1/3) = 10, fyy (1/3,�1/3) = fxy (1/3,�1/3) = 2, andfxx fyy > (fxy )

2, so (1/3,�1/3) is a relative minimum point.


Quadratic Forms

A function q with n variables is said to have the quadratic form if itcan be written as

q (u1, � � �, un) = d11u21 + 2d12u1u2 + � � �+ 2d1nu1un (55)

+d22u22 + 2d23u2u3 + � � �+ 2d2nu2un+ � � �+dnnu2n

If we let dij = dji , i < j , then q (u1, � � �, un) can be written as

q (u1, � � �, un) =n

∑i=1

n

∑j=1dijuiuj = u0Du, (56)

where the quadratic-form systematic matrix

D =

24 d11 � � � d1n� � � � � � � � �dn1 � � � dnn

35 and u0 =�u1 � � � un

�(57)


De�nitionA quadratic form q (u1, � � �, un) = u0Du is said to be (a) positive de�niteif q (u) > 0 if all u 6= 0; (b) positive semi-de�nite if q (u) � 0 if all u 6= 0;(c) negative de�nite if q (u) < 0 if all u 6= 0; (d) negative semi-de�nite ifq (u) � 0 if all u 6= 0.

Sometimes we say that a matrix D is positive (negative) de�nite ifthe corresponding quadratic form is positive (negative) de�nite.


Determinantal Test for Sign De�niteness

TheoremFor the quadratic form q (u) = u0Du, the necessary and su¢ cientconditions for positive de�niteness is that the principle minors of jD j arepositive:

jD1j = d11 > 0, jD2j =�� d11 d12d21 d22

�� > 0, � � �,Dn =24 d11 � � � d1n� � � � � � � � �dn1 � � � dnn

35 > 0.The corresponding N&S condition for negative de�niteness is that thePMs alternates in sign:

jD1j < 0, jD2j > 0, jD3j < 0, � � �. (58)


(conti.) Two-variable quadratic-form case: Consider the second-ordertotal di¤erential

d2z = fxxdx2 + 2fxydxdy + fyydy2

=�dx dy

� � fxx fxyfxy fyy

�dxdy

. (59)

Thus, for the function z = f (x , y) , (a) d2z is positive de�nite i¤fxx > 0 and fxx fyy > (fxy )

2 ; (b) d2z is negative de�nite i¤ fxx < 0and fxx fyy > (fxy )

2 . Note that since fxx fyy � (fxy )2 > 0, fxx and fyymust have the same sign. Hence, this is just the SO su¢ cientcondition presented before.

The determinant jH j =�� fxx fxyfxy fyy

�� is referred to as the discriminantof the quadratic-form of f and is also called the Hessian determinant.


(conti.) Example: q = 5u2 + 3uv + 2v2 is positive de�nite because

the discriminant of q is

�� 10 33 4

�� , with leading principle minors:10 > 0 and

�� 10 33 4

�� = 31 > 0.Example: Given that fxx = �2, fyy = �1, and fxy = 1 at a certainpoint on a function z = f (x , y) , does d2z have a de�nite sign atthat point regardless of the values of dx and dy? In this case, d2z isnegative de�nite because the discriminant of the quadratic-form d2z

is in the case

�� 2 11 �1

�� , with leading PMs�2 < 0 and

�� 2 11 �1

�� = 1 > 0.Luo, Y. (SEF of HKU) MME January 21, 2009 37 / 46

Objective Functions with More than Two Variables

Consider the following function z = f (x1, x2, � � �, xn) ,the totaldi¤erential is

dz = f1dx1 + f2dx2 + � � �+ fndxn, (60)

so that the necessary condition for extremum is dz = 0 for arbitrarydxi , which in turn means that all n FO partial derivatives are 0 :

f1 = f2 = � � � = fn = 0. (61)

It can be veri�ed that the SO di¤erential d2z is

d2z =�dx1 � � � dxn

� 24 f11 � � � f1n� � � � � � � � �fn1 � � � fnn

3524 dx1� � �dxn

35= (dx)T Hdx (62)

Thus the Hessian determinant is jH j and the SO su¢ cient conditionfor extremum is that all n principal minors be positive for a minimumin z and that they duly alternate in sign for a maximum, the �rst onebeing negative.Luo, Y. (SEF of HKU) MME January 21, 2009 38 / 46

TheoremConditions for maximum: (1) f1 = f2 = � � � = fn = 0 (necessarycondition); (2) jH1j < 0, jH2j > 0, jH3j < 0, � � �, (�1)n jHn j > 0. d2z isnegative de�nite.

TheoremConditions for minimum: (1) f1 = f2 = � � � = fn = 0 (necessarycondition); (2) jH1j > 0, jH2j > 0, � � �, jHn j > 0. d2z is negative de�nite.


(conti.) Example: Find the extreme values ofz = 2x21 + x1x2 + 4x

22 + x1x3 + x

23 + 2. Step 1: Deriving the FOCs:

f1 = f2 = f3 = 0 =) x�1 = x�2 = x

�3 = 0,

which means that there is only one stationary value z� = 2. Step 2:Deriving the SO partial derivatives and forming the Hessiandeterminant:

jH j =

��f11 f12 f13f21 f22 f23f31 f32 f33

�� =��4 1 11 8 01 0 2

�� ,with jH1j = 4 > 0, jH2j = 31 > 0, jH3j = 54 > 0. Hence, z� = 2 is aminimum.


SO conditions and Concavity&Convexity

SO conditions are always concerned with whether a stationary point isthe peak of a hill or the bottom of a valley and are closely related tothe (strictly) concave or convex functions.

De�nitionsA function that gives rise to a hill (valley) over the entire domain is said tobe a concave (convex) function. If the hill (valley) pertains only to asubset S of the domain, the function is said to be concave (convex) on S .

De�nitionsMathematically, the function f is concave (convex) i¤ for any pair ofdistinct points u and v in the domain of f , for any θ 2 (0, 1),

θf (u) + (1� θ) f (v) � (�) f (θu + (1� θ) v) . (63)

Further, if �� (or �)� is replaced by �< (or >)�the function is said tobe strictly concave (strictly convex).


Theorem (Linear functions)

If f (x) is a linear function, then it is a concave function as well as aconvex function, but not strictly.

Theorem (Negative of a function)

If f (x) is a (strictly) concave function, then �f is a (strictly) convexfunction, and vice versa.

Theorem (Sum of functions)

If f (x) and g (x) are both concave (convex) functions, then f (x) + g (x)is also a concave (convex) function. Further, if either one or both arestrictly concave (strictly convex), the f (x) + g (x) is strictly concave(strictly convex).


FactIn view of the association of concavity (convexity) with a global hill(valley) con�guration, an extremum of a concave (convex) function mustbe a peak-a maximum (a bottom - a minimum). Moreover, the maximum(minimum) must be an absolute maximum (minimum). Further, themaximum (minimum) is unique if the function is strictly concave (strictlyconvex).

FactNote that here the properties of concavity or convexity are taken to beglobal in scope. If they are valid only in a subset S , the associatedmaximum or minimum are relative to that subset.


TheoremA twice continuously di¤erentiable function z = f (x1, x2, � � �, xn) isconcave (convex) i¤ d2z is everywhere negative (positive) semide�nite.The said function is strictly concave (convex) if (but not only if) d2z iseverywhere negative (positive) de�nite.

Hence, we can easily verify whether a function is strictly concave(strictly convex) by checking whether its Hessian matrix is negative(positive) de�nite.

Example: For z = x21 + x22 , since

jH j =�� f11 f12f21 f22

�� = �� 2 00 2

�� ,with jH1j = 2 > 0, jH2j = 4 > 0, the function is strictly convex.


Firm�s Optimal Investment Decision Problem

Consider a competitive �rm with the following pro�t function

π = R � C = PQ � wL� rK (64)

where P is the price of output, Q is the quantity of output, w and r aretwo factor-prices, K and L are two factors: capital and labor, respectively.

Since the market is competitive, the �rm cannot a¤ect the prices andthus P, w and r are given exogenously. Q, K and L are threeendogenous variables in the model.Assume that Q is a function of K and L via a Cobb-Douglasproduction function

Q = Q (K , L) = LαK β.

For simplicity, we assume that α = β < 12 .

Substituting it into the pro�t function gives

π (L,K ) = PLαK α � wL� rK (65)


FOCs for pro�t maximization are

∂π

∂L= αPLα�1K α � w = 0 (66)

∂π

∂K= αPLαK α�1 � r = 0, (67)

which de�nes the optimal levels of K and L that maximizes pro�t.

Second, let�s check the SO condition to verify that we do have amaximum. The Hessian for this problem is

jH j =�� πLL πLK

πKL πKK

�� = �� α (α� 1)PLα�2K α α2PLα�1K α�1

α2PLα�1K α�1 α (α� 1)PLαK α�2

�� .Hence, the su¢ cient condition for a maximum is that

jH1j = α (α� 1)PLα�2K α < 0, jH2j = (1� 2α)α2P2L2α�2K 2α�2 > 0,

which means that when α < 1/2, the SO su¢ cient condition issatis�ed. Combining the two FOCs can easily solve for K � and L�.


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