1

4. Models with MultipleExplanatory Variables

Chapter 2 assumed that the dependent variable (Y) is affected by only ONE explanatory variable (X).

Sometimes this is the case. Example: Age = Days Alive/365.25

Usually, this is not the case. Example: midterm mark depends on: how much you study how well you study intelligence, etc

2

4. Multi Variable Examples:Demand = f( price of good, price of substitutes,

income, price of compliments)

Consumption = f( income, tastes, wages)

Graduation rates = f( tuition, school quality, student quality)

Christmas present satisfaction = f (cost, timing, knowledge of person, presence of card, age, etc.)

3

4. The Partial DerivativeIt is often impossible analyze ONE variable’s

impact if ALL variables are changing.

Instead, we analyze one variable’s impact, assuming ALL OTHER VARIABLES REMAIN CONSTANT

We do this through the partial derivative.

This chapter uses the partial derivative to expand the topics introduced in chapter 2.

4

4. Calculus and Applications involving More than One Variable

4.1 Derivatives of Functions of More Than One Variable

4.2 Applications Using Partial Derivatives4.3 Partial and Total Derivatives4.4 Unconstrained Optimization4.5 Constrained Optimization

5

4.1 Partial DerivativesConsider the function z=f(x,y). As this function

takes into account 3 variables, it must be graphed on a 3-dimensional graph.

A partial derivative calculates the slope of a 2-dimensional “slice” of this 3-dimensional graph.

The partial derivative ∂z/∂x asks how x affects z while y is held constant (ceteris paribus).

6

4.1 Partial DerivativesIn taking the partial derivative, all other variables

are kept constant and hence treated as constants (the derivative of a constant is 0).

There are a variety of ways to indicate the partial derivative:

1) ∂y/∂x2) ∂f(x,z)/∂x3) fx(x,z)Note: dy=dx is equivalent to ∂y/∂x if y=f(x); ie: if y

only has x as an explanatory variable.(Therefore often these are used interchangeably

in economic shorthand)

7

4.1 Partial DerivativesLet y = 2x2+3xz+8z2

∂y/ ∂x = 4x+3z+0∂y/ ∂z = 0+3x+16z

(0’s are dropped)Let y = xln(zx)∂ y/ ∂ x = ln(zx) + zx/zx

= ln(zx) + 1∂ y/ ∂ z = x(1/zx)x

=x/z

8

4.1 Partial DerivativesLet y = 3x2z+xz3-3z/x2

∂ y/ ∂ z=3x2+3xz2-3/x2

∂ y/ ∂ x=6xz+z3+6z/x3

Try these:

z=ln(2y+x3)Expenses=sin(a2-ab)+cos(b2-ab)

9

4.1.1 Higher Partial DerivativesHigher order partial derivates are evaluated

exactly like normal higher order derivatives.It is important, however, to note what variable to

differentiate with respect to:

From before:Let y = 3x2z+xz3-3z/x2

∂ y/ ∂ z=3x2+3xz2-3/x2

∂ 2y/ ∂ z2=6xz∂ 2y/ ∂ z ∂ x=6x+3z2+6/x3

10

4.1.1 Young’s TheoremFrom before:Let y = 3x2z+xz3-3z/x2

∂ y/ ∂ x=6xz+z3+6z/x3

∂ 2y/ ∂ x2=6z-18z/x4

∂ 2y/ ∂ x ∂ z=6x+3z2+6/x3

Notice that ∂2y/∂x∂z=∂2y/∂z∂xThis is reflected by YOUNG’S THEOREM:

order of differentiation doesn’t matter for higher order partial derivatives

11

4.2 Applications using Partial Derivatives

As many real-world situations involve many variables, Partial Derivatives can be used to analyze our world, using tools including:

Interpreting coefficients Marginal Products

12

4.2.1 Interpreting CoefficientsGiven a function a=f(b,c,d), the dependent

variable a is determined by a variety of explanatory variables b, c, and d.

If all dependent variables change at once, it is hard to determine if one dependent variables has a positive or negative effect on a.

A partial derivative, such as ∂ a/ ∂ c, asks how one explanatory variable (c), affects the dependent variable, a, HOLDING ALL OTHER DEPENDENT VARIABLES CONSTANT (ceteris paribus)

13

4.2.1 Interpreting CoefficientsA second derivative with respect to the same

variable discusses curvature.A second cross partial derivative asks how the

impact of one explanatory variable changes as another explanatory variable changes.

Ie: If Happiness = f(food, tv),∂ 2h/ ∂ f ∂tv asks how watching more tv affects

food’s effect on happiness (or how food affects tv’s effect on happiness). For example, watching TV may not increase happiness if someone is hungry.

14

4.2.1 Corn ExampleConsider the following formula for corn

production:

Corn = 500+100Rain-Rain2+50Scare*FertilizerCorn = bushels of cornRain = centimeters of rainScare=number of scarecrowsFertilizer = tonnes of fertilizer

Explain this formula

15

4.2.1 Corny Example1) Intercept = 500

-if it doesn’t rain, there are no scarecrows and no fertilizer, the farmer will harvest 500 bushels

2) ∂Corn/∂Rain=100-2Rain-each additional cm of rain changes corn production by 100-2Rain-positive impact if rain < 50 cm-negative impact if rain > 50 cm

16

4.2.1 Corny Example

3) ∂2Corn/∂Rain2=-2<0, (concave)

-More rain has a DECREASING impact on the corn harvest-More rain DECREASES rain’s impact on the corn harvest by 2

4) ∂Corn/∂Scare=50Fertilizer

-More scarecrows will increase the harvest 50 for every tonne of fertilizer-if no fertilizer is used, scarecrows are useless

17

4.2.1 Corny Example5) ∂ 2Corn/∂Scare2=0 (straight line, no curvature)

-Additional scarecrows have a CONSTANT impact on corn’s harvest

6) ∂ 2Corn/∂Scare∂Fertilizer=50-Additional fertilizer increases scarecrow’s impact on the corn harvest by 50

18

4.2.1 Corny Example

7) ∂Corn/∂Fertilizer=50Scare

-More fertilizer will increase the harvest 50 for every scarecrow-if no scarecrows are used, fertilizer is useless

8) ∂ 2Corn/∂Fertilizer2=0, (straight line)

-Additional fertilizer has a CONSTANT impact on corn’s harvest

19

4.2.1 Corny Example9) ∂ 2Corn/∂Fertilizer ∂Scare =50

-Additional scarecrows increase fertilizer’s impact on the corn harvest by 50

20

4.2.2 Partial Derivatives and Marginal Product

Consider the function Q=f(L,K,)

The partial derivative is the change in output if one input (labour or capital) increases by one.

The partial derivative IS the MARGINAL PRODUCT of an production function

∂ Q/ ∂ L=Marginal Product of Labour (MPL)∂ Q/ ∂ K=Marginal Product of Capital (MPK)

21

4.2.2 Cobb-Douglas Production Function

A favorite function of economists is the Cobb-Douglas Production Function of the form

Q=aLbKcOf

Where L=labour, K=Capital, and O=Other (education, technology, government, etc.)

This is an attractive function because if b+c+f=1, the demand function is homogeneous of degree 1. (Doubling all inputs doubles outputs…a happy concept)

22

4.2.2 Cobb-Douglas UniversityConsider a production function for university

degrees:

Q=aLbKcAf

Where

L=Labour (ie: professors), K=Capital (ie: classrooms)A=Administration

23

4.2.2 Average and Marginal ProductsFinding partial derivatives:∂ Q/ ∂ L =abLb-1KcAf

=b(aLbKcAf)/L=b(Q/L) OR=b* average product of labour

-in other words, adding an additional professor will contribute a fraction of the average product of each current professor

-this partial derivative gives us the MARGINAL PRODUCT of labour

24

4.2.2 Cobb-Douglas ProfessorsFor example, if 20 professors are employed by

the department, and 500 students graduate yearly, and b=0.5:

∂ Q/ ∂ L =0.5(500/20)=12.5

Ie: Hiring another professor will graduate 12.5 more students. The marginal product of labour is 12.5

25

4.2.2 Partial ElasticitiesIn chapter 2, when y=f(x), elasticities were

calculated as:

yx

dxdy

And calculated the % change in y when x increases by 1%

When y=f(x,z), PARTIAL elasticities are calculated as:

yz

zy

yx

xy

zyxy

,, or

And calculated the % change in y when only x (or only y) increases by 1%.

26

4.2.2 Partial Elasticity ExampleFred’s Fantastic Fans has income (I) that

depends on price (P) and advertising (A) according to:

2PPAI

If price=$100 and advertising=$1,000, calculate the % change in income if Fred increases advertising by 1%.

27

4.2.2 Partial Elasticity Example

%11.1000,90000,100

100)000,1(100)000,1(100)000,1,100(

PAPA

,

2,

2,

AI

AI

AI

AP

PPAIIA

AI

If Fred increases advertising by 1%, his income would increase by 1.11%.

28

4.2.2 Partial ElasticitiesIn chapter 2, we saw that elasticities can be

calculated using logs:

xdyd

lnln

When y=f(x,z), PARTIAL elasticities can also be calculated using logs:

zyxy

zy

xy

lnln

lnln

,

,

29

4.2.2 Logs and CobbsWe can highlight elasticities by using logs:

Q=aLbKcCf

Using log rules, this converts to:

Ln(Q)=ln(a)+bln(L)+cln(k)+fln(C)

We now find that:εQ,L= ∂ ln(Q)/ ∂ ln(L)=b

Using logs, elasticities more apparent.

30

4.2.2 Logs and Demand

Consider a log-log demand example:

Ln(Qdx)=ln(β1) +β2 ln(Px)+ β3 ln(Py)+ β4 ln(I)

We now find that:Own Price Elasticity = β2

Cross-Price Elasticity = β3

Income Elasticity = β4

31

4.2.2 ilogsConsidering the demand for the ipad, assume:

Ln(Qdipad)=2.7 -1ln(Pipad)+4 ln(Ptablet)+0.1 ln(I)

We now find that:Own Price Elasticity = -1, demand is unit elasticCross-Price Elasticity = 4, a 1% increase in the

price of tablets causes a 4% increase in quantity demanded of ipads

Income Elasticity = 0.1, a 1% increase in income causes a 0.1% increase in quantity demanded for ipads

32

4.3 Total Derivatives

Often in econometrics, one variable is influenced by a variety of other variables.

Ie: Happiness =f(sun, driving)Ie: Productivity = f(labor, effectiveness)

Using TOTAL DERIVATIVES, we can examine how growth of one variable is caused by growth in all other variables

The following formulae will combine x’s impact on y (dy/dx) with x’s impact on y, with other variables held constant (δy/δx)

Assume you are increasing the square footage of a house where

AREA = LENGTH X WIDTHA=LWIf you increase the length, the change in area is equalto the increase in length times the current width:

33

4.3 Total Derivatives

Length

Area

Width

dL

Notice that:δA/δL=W, (partial derivative, since width is constant)Therefore the increase in area is equal to:dA=(δA/δL)dL

A=LWIf you increase the width, the change in area is equalto the increase in width times the current length:

34

4.3 Total DerivativesLength

Area

Width

dW

Notice that:δA/δW=L, (partial derivative, since length is constant)Therefore the increase in area is equal to:dA=(δA/δW)dW

Next we combine the two effects:

A=LWAn increase in both lengthand width has the followingimpact on area:

35

4.3 Total DerivativesLength

Area

Width

dW

Now we have:dA=(δA/δL)dL+(δA/δW)dW+(dW)dL

But since derivatives always deal with instantaneous slope and small changes, (dW)dL is small and ignored, resulting in:

dA=(δA/δL)dL+(δA/δW)dW

dL

36

4.3 Total DerivativesLength

Area

Width

dW

dA=(δA/δL)dL+(δA/δW)dW

Effectively, we see that change in the dependent variable (A), comes from changes in the independent variables (W and L). In general, given the function z=f(x,y) we have:

dL

dyyzdx

xzdy

yyxfdx

xyxfdz

),(),(

In a joke factory,

QJokes=workers(funniness)

You employ 500 workers, each of which can create 100 funny jokes an hour.

How many more jokes could you create if you increase workers by 2 and their average funniness by 1 (perhaps by discovering any joke with an elephant in it is slightly more funny)? 37

4.3 Total Derivative Example

700200500)2(100)1(500

dqdq

fdwwdfdq

dwwqdf

fqdq

The key advantage of the total derivative is it takes variable interaction into account.

The partial derivative (δz/δx) examines the effect of x on z if y doesn’t change. This is the DIRECT EFFECT.

However, if x affects y which then affects z, we might want to measure this INDIRECT EFFECT.

We can modify the total derivative to do this:

38

4.3 Total Derivative Extension

dxdyyz

xz

dxdyyz

dxdxxz

dxdz

dyyzdx

xzdy

yyxfdx

xyxfdz

),(),(

39

4.3 Total Derivative Extension

dxdyyz

xz

dxdyyz

dxdxxz

dxdz

dyyzdx

xzdy

yyxfdx

xyxfdz

),(),(

Here we see that x’s total impact on z is broken up into two parts:

1) x’s DIRECT impact on z (through the partial derivative)

2) x’s INDIRECT impact on z (through y)

Obviously, if x and y are unrelated, (δy/δx)=0, then the total derivative collapses to the partial derivative

40

4.3 Total Derivative ExampleAssume Happiness=Candy+3(Candy)Money+Money2

h=c+3cm+m2

Furthermore, Candy=3+Money/4 (c=3+m/4)

The total derivative of happiness with regards to money:

mcdmdh

mmcdmdh

dmdc

ch

mh

dmdh

75.2325.0

)]4/1)(31[(]23[

41

4.3 Total Derivative and ElasticityTotal derivatives can also give us the relationship between elasticity and revenue that we found in Chapter 2.2.3:

demand) of elasticity price is (where )1(

)1(

QdPdTR

dPdQ

QPQ

dPdTR

dPdQP

dPdPQ

dPdTR

dQQTRdP

PTRdTR

PQTR

42

4.4 Unconstrained Optimization

Unconstrained optimization falls into two categories:

1) Optimization using one variable (ie: changing wage to increase productivity, working conditions are constant)

2) Optimization using two (or more) variables (ie: changing wage and working conditions to maximize productivity)

43

4.4 Simple Unconstrained OptimizationFor a multivariable case where only one variable is

controlled, optimization steps are easy:Consider the function z=f(x)1) FOC:Determine where δz/δx=0 (necessary condition)2) SOC:δ2z/δx2<0 is necessary for a maximumδ2z/δx2>0 is necessary for a minimum3) Determine max/min pointSubstitute the point in (2) back into the original

equation.

44

4.4 Simple Unconstrained OptimizationLet productivity = -wage2+10wage(working conditions)2

P(w,c)=-w2+10wc2 If working conditions=2, find the wage that maximizes

productivityP(w,c)=-w2+40w

1) FOC:δp/δw =-2w+40=0 w=202) SOC:δ2p/δw2= -2 < 0, a maximum exists

45

4.4 Simple Unconstrained OptimizationP(w,c)=-w2+10wc2 w=20 (maximum confirmed)

3) Find MaximumP(20,4)=-202+10(20)(2)2

P(20,4)=-400+800P(20,4)=400

Productivity is maximized at 400 when wage is 20.

46

4.4 Complex Unconstrained OptimizationFor a multivariable case where only two variable are

controlled, optimization steps are more in-depth:Consider the function z=f(x,y)

1) FOC:Determine where δz/δx=0 (necessary condition)AndDetermine where δz/δy=0 (necessary condition)

47

4.4 Complex Unconstrained OptimizationFor a multivariable case where only two variable are

controlled, optimization steps harder:Consider the function z=f(x,y)2) SOC:δ2z/δx2<0 and δ2z/δy2<0 are necessary for a maximumδ2z/δx2>0 and δ2z/δy2>0 are necessary for a minimumPlus, the cross derivatives can’t be too large compared

to the own second partial derivatives:

022

2

2

2

2

yxz

yz

xz

48

4.4 Complex Unconstrained Optimization

If this third SOC requirement is not fulfilled, a SADDLE POINT occurs, where z is a maximum with regards to one variable but a minimum with regards to the other. (ie: wage maximizes productivity while working conditions minimizes it)

Vaguely, even though both variables work to increase z, their interaction with each other outweighs this maximizing effect

022

2

2

2

2

yxz

yz

xz

49

4.4 Complex Unconstrained OptimizationLet P(w,c)=-w2+wc-c2 +9c , maximize productivity1) FOC:δp/δw =-2w+c=0 2w=cδp/δc=w-2c+9=0w=2c-9w=2(2w)-9-3w=-9w=32w=c6=c

50

4.4 Complex Unconstrained OptimizationP(w,c)=-w2+wc-c2 +9c δp/δw =-2w+c=0 δp/δc=w-2c+9=0w=3, c=6 (possible max/min)2) SOC:δ2p/δw2= -2 < 0δ2p/δc2= -2 < 0, possible max

03

1)2)(2(

22

2

2

2

2

222

2

2

2

2

cwp

cp

wp

cwp

cp

wp Maximum confirmed

51

4.4 Complex Unconstrained OptimizationP(w,c)=-w2+wc-c2 +9c w=3, c=6 (confirmed max)3) Find productivity:

27),(

5436189),()6(966)3(3),(

9),(22

22

cwpcwpcwp

ccwcwcwp

Productivity is maximized at 27 when wage=3 and working conditions=6.

52

4.5 Constrained OptimizationTypically constrained optimization consists of

maximizing or minimizing an objective function with regards to a constraint, or

Max/min z=f(x,y)Subject to (s.t.): g(x,y)=k

Where k is a constant

53

4.5 Constrained OptimizationOften economic agents are not free to make any

decision they would like. They are CONSTRAINED by factors such as income, time, intelligence, etc.

When optimizing with constraints, we have two general methods:

1) Internalizing the constraint2) Creating a Lagrangeian function

54

4.5 Internalizing ConstraintsIf the constraint can be substituted into the equation to

be optimized, we are left with an unconstrained optimization problem:

Example:Bob works a full week, but every Saturday he has

seven hours left free, either to watch TV or read. He faces the constrained optimization problem:

Max. Utility=7TV-TV2+Read (U=7TV-TV2+R)s.t. 7=TV+Read (7=TV+R)

55

4.5 Internalizing ConstraintsMax. U=7TV-TV2+R

s.t. 7=TV+R

We can solve the constraint:R=7-TV

And substitute into the objective function:U=-TV2+7TV+(7-TV)U=-TV2+6TV+7

56

4.5 Internalizing ConstraintsMax. U=7TV-TV2+R

s.t. 7=TV+RU=-TV2+6TV+7We can then perform unconstrained optimization:FOC:δU/ δTV=-2TV+6=0TV=3R=7-TVR=7-3R=4

57

4.5 Internalizing ConstraintsMax. U=7TV-TV2+R

s.t. 7=TV+RU=-TV2+6TV+7, TV=3, R= 4δU/ δTV=-2TV+6SOC:δ2U/ δTV2=-2<0, concave max.

Evaluate:U=7TV-TV2+RU=7(3)-32+4U=21-9+4=16

58

4.5 Internalizing ConstraintsMax. U=7TV-TV2+R

s.t. 7=TV+RU=-TV2+6TV+7, TV=3, R= 4δU/ δTV=-2TV+6δ2U/δTV2=-2<0, concave max.U=21-9+4=16

Utility is maximized at 16 when Bob watches 3 hours of TV and reads for 4 hours.

59

4.5 Internalizing ConstraintsSubstituting the constraint into the objective function

may not be applicable for a variety of reasons:1) The substitution makes the objective function unduly

complicated, or substitution is impossible2) You want to evaluate the impact of the constraint3) The constraint is an inequality4) Your exam paper asks you to do so

In this case, you must construct a Lagrangian function.

60

4.5 The LagrangianGiven the optimization problem:

Max/min z=f(x,y)s.t. g(x,y)=k (Where k is a constant)

The Lagranean (Lagrangian) function becomes:

L=z*=z(x,y)+λ(k-g(x,y))

Where λ is known as the Lagrange Multiplier.

We then continue with FOC’s and SOC’s.

61

4.5 The LagrangianL=z*=z(x,y)+λ(k-g(x,y))

FOC’s:0 ,0 ,0

LLL

yxNote that the third FOC simply returns the constraint,

g(x,y)=kTypically, one will solve for λ in the first two conditions

to find a relationship between x and y, then use this relationship with the third condition to solve for x and y.

62

4.5 The LagrangianL=z*=z(x,y)+λ(k-g(x,y))

After finding FOC’s, to confirm a maximum or minimum, the SOC is employed.

This SOC must be negative for a maximum and positive for a minimum

Note that for more terms, this function becomes exponentially complicated.

SOC’s:

δxδyzδ

δyδg

δxδg2

δxzδ

δyδg

δyzδ

δxδg 2

2

22

2

22

SOC

63

4.5 Lagrangian exampleMax. U=7TV-TV2+R

s.t. 7=TV+R

L=z*=z(x,y)+λ(k-g(x,y))L=7TV-TV2+R+λ(7-TV-R)FOC:

2TV-7

0-2TV-7

TVL

1

0-1

RL

RV

T

T7

0R-V-7L

64

4.5 Lagrangian example

RTV

7)3(1)2(

2TV-7)1(

3127

)2()1(

TVTV

RRRV

437T7:)3(

65

4.5 Lagrangian Example

02

01122-101

δTVδRzδ

δRδg

δTVδg2

δTVzδ

δRδg

δRzδ

δTVδg

22

2

2

22

2

22

SOCSOC

SOC

Since the second order condition is negative, the points found are a maximum.

Notice that we found the same answers as internalizing the constraint.

66

4.5 The Lagrange MultiplierThe Lagrange Multiplier, λ, provides a measure of how

much of an impact relaxing the constraint would make, or how the objective function changes if k of g(x,y)=k is marginally increased.

The Lagrange multiplier answers how much the maximum or minimum changes when the constraint g(x,y)=k increases slightly to g(x,y)=k+δ

67

4.5 Lagrangian example

12(3)-7 2TV-74

3 2TV-7

RTV

This means that if Bob gets an extra hour, his maximum utility will increase by approximately 1.

(Alternately, if Bob loses an hour of leisure, his maximum utility will decrease by approximately 1.)

Check: If 8=TV+R, TV=3.5, R=4.5, U=16.75

(utility increases by approximately 1)