1

Technology

Beattie, Taylor, and WattsSections: 2.1-2.2, 5.1-5.2b

2

Agenda The Production Function with One

Input Understand APP and MPP Diminishing Marginal Returns and

the Stages of Production The Production Function with Two

Input Isoquants

3

Agenda Cont. Marginal Rate of Technical

Substitution Returns to Scale Production Possibility Frontier Marginal Rate of Product

Transformation

4

Production Function A production function maps a set of

inputs into a set of outputs. The production function tells you how to

achieve the highest level of outputs given a certain set of inputs.

Inputs to the production function are also called the factors of production.

The general production function can be represented as y = f(x1, x2, …, xn).

5

Production Function Cont. The general production function

can be represented as y = f(x1, x2, …, xn). Where y is the output produced and is

a positive number. Where xi is the quantity of input i for i

= 1, 2, …, n and each is a positive number.

6

Production Function with One Input In many situations, we want to

examine what happens to output when we only change one input. This is equivalent to investigating the

general production function previously given holding all but one of the variables constant.

7

Production Function with One Input Cont. Mathematically we can represent

the production function with one input as the following: y = f(x) = f(x1) = f(x1|x2,x3,…,xn) Suppose y = f(x1, x2, x3) = x1*x2*x3 Suppose that x2 = 3 and x3 = 4, which

are fixed inputs, then y = f(x) = f(x1) = f(x1|3,4) = 12x1 =12x

8

Example of Production Function y = f(x) = -x3 + 60x2

Production Function

0

5000

10000

15000

20000

25000

30000

35000

0 10 20 30 40 50 60

y

9

APP and MPP There are two major tools for

examining a production function: Average Physical Product (APP) Marginal Physical Product (MPP)

10

APP The average physical product tells

you the average amount of output you are getting for an input.

We define APP as output (y) divided by input (x). APP = y/x = f(x)/x

11

Example of Finding APP Assume you have the following

production function: y = f(x) = -x3 + 60x2

)60(

60

60

2

23

xxAPP

xxAPP

x

xx

x

yAPP

12

Example of Finding Maximum APP To find the maximum APP, you take the

derivative of APP and solve for the x that gives you zero.

From the previous example: APP = -x2 + 60x

30

0602

60max 2

x

xdx

dAPP

xxdx

d

dx

dAPPAPP

13

MPP The marginal physical product tells you

what effect a change of the input will do to the output. In essence, it is the change in the output

divided by the change in the input. MPP is defined as:

)(' xfdx

dyMPP

14

Example of Finding MPP Assume you have the following

production function: y = f(x) = -x3 + 60x2

)40(3

1203

60

2

23

xxMPP

xxMPP

xxdx

d

dx

dyMPP

15

Interpreting MPP When MPP > 0, then the production

function is said to have positive returns to the use of the input. This occurs on the convex and the

beginning of the concave portion of the production function.

In the previous example, this implies that MPP > 0 when input is less than 40 (x<40).

16

Interpreting MPP Cont. When MPP = 0, then we know that

the production function is at a maximum. Setting MPP = 0 is just the first order

condition to find the maximum of the production function.

In the example above, MPP = 0 when the input was at 40.

17

Interpreting MPP Cont. When MPP < 0, then the production

function is said to have decreasing returns to the use of the input. This occurs on the concave portion of

the production function. In the previous example, this implies

that MPP < 0 when input is greater than 40 (x>40).

18

Example of APP and MPP y = f(x) = -x3 + 60x2

APP and MPP

-2000

-1500

-1000

-500

0

500

1000

1500

0 10 20 30 40 50 60APP

MPP

19

Law of Diminishing Marginal Returns (LDMR) The Law of Diminishing Marginal

Returns states that as you add successive units of an input while holding all other inputs constant, then the marginal physical product must eventually decrease. This is equivalent to saying that the

derivative of MPP is negative.

20

Finding Where LDMR Exists Suppose y = f(x) = -x3 + 60x2

To find where the LDMR exists is equivalent to finding what input levels give a second order condition that is negative.

20

0)20(0)20(60

negative termabove set the exists LDMR wherefind To

)20(61206

1203

2

2

2

x

xxdx

dMPP

xxdx

yd

dx

dMPP

xxdx

dyMPP

21

Relationship of APP and MPP When MPP > APP, then APP is

rising When MPP = APP, then APP is at a

maximum When MPP < APP, then APP is

declining

22

Relationship of APP and MPP Cont.

APPMPPdx

dAPPx

xfxf

dx

dAPP

xfxxfdx

dAPP

xxfxxfdx

dAPP

x

xfxxf

dx

dAPP

x

xfxxf

dx

dAPP

x

xf

dx

d

dx

dAPPx

xfAPP

when 0

)()(' when 0

)()(' when 0

*0)()(' when 0

0)()('

when 0

)()('

)(

)(

2

2

2

23

Stages of Production Stage I of production is where the

MPP is above the APP, i.e., it starts where the input level is 0 and goes all the way up to the input level where MPP=APP. To find the transition point from stage I

to Stage II you need to set the APP function equal to the MPP function and solve for x.

24

Stages of Production Cont. Stage II of production is where MPP is

less than APP but greater than zero, i.e., it starts at the input level where MPP=APP and ends at the input level where MPP=0. To find the transition point from Stage II to

Stage III, you want to set MPP = 0 and solve for x.

Stage III is where the MPP<0, i.e., it starts at the input level where MPP=0.

25

Graphical View of the Production Stages

Stage II

Stage I

Stage III

TPP

Y

xMPP APP

xMPP

APP

26

Finding the Transition From Stage I to Stage II Suppose y = f(x) = -x3 + 60x2

30or 0

0)30(2

0602

120360

for x solve and APP MPPset point n transitio thefind To

601203

1203

2

22

22

2

xx

xx

xx

xxxx

MPPAPP

xxx

xx

x

yAPP

xxdx

dyMPP

27

Finding the Transition From Stage II to Stage III Suppose y = f(x) = -x3 + 60x2

40or 0

0)40(3

01203

0

for x solve and 0 MPPset point n transitio thefind To

1203

2

2

xx

xx

xx

MPP

xxdx

dyMPP

28

Production Function with Two Inputs While one input production

functions provide much intuitive information about production, there are times when we want to examine what is the relationship of output to two inputs. This is equivalent to investigating the

general production function holding all but two of the variables constant.

29

Production Function with Two Inputs Cont. Mathematically we can represent

the production function with one input as the following: y = f(x1,x2) = f(x1, x2|x3,…,xn)

30

Example of a Production Function with Two Variables: y=f(x1,x2)=-x1

3+25x12-

x23+25x2

2

0 2 4 6 8

10 12 14 16 18 20 22 240

6

12

18

24

0.00

500.00

1000.00

1500.00

2000.00

2500.00

3000.00

3500.00

4000.00

4500.00

5000.00

4500.00-5000.00

4000.00-4500.00

3500.00-4000.00

3000.00-3500.00

2500.00-3000.00

2000.00-2500.00

1500.00-2000.00

1000.00-1500.00

500.00-1000.00

0.00-500.00

31

Example 2 of a Production Function with Two Variables: y=f(x1,x2)=8x1

1/4x23/4

0

2

4

6

8 10

012345678910

0

10

20

30

40

50

60

70

80

70-80

60-70

50-60

40-50

30-40

20-30

10-20

0-10

32

Three Important Concepts for Examining Production Function with Two Inputs

There are three important concepts to understand with a production function with two or more inputs. Marginal Physical Product (MPP) Isoquant Marginal Rate Of Technical

Substitution (MRTS)

33

MPP for Two Input Production Function MPP for a production function with

multiple inputs can be viewed much like MPP for a production function with one input. The only difference is that the MPP for the

multiple input production function must be calculated while holding all other inputs constant, i.e., instead of taking the derivative of the function, you take the partial derivative.

34

MPP for Two Input Production Function Cont. Hence, with two inputs, you need to

calculate the MPP for both inputs. MPP for input xi is defined

mathematically as the following:

22

11

2

21

1

21

21

),(

),(

),(

xx

xx

xi

x

fx

xxfMPP

fx

xxfMPP

fx

xxfMPP

ii

35

Example of Calculating MPP Suppose y = f(x1,x2) = -x1

3+25x12-x2

3+25x22

222

2

21

121

1

21

22

32

21

3121

503),(

503),(

2525),(

2

1

xxx

xxfMPP

xxx

xxfMPP

xxxxxxfy

x

x

36

Example 2 of Calculating MPP Suppose y = f(x1,x2) = 8x1

1/4 x23/4

4

1

2

1

4

1

2

4

1

14

1

24

1

14

1

24

1

12

21

4

3

1

2

4

3

1

4

3

24

3

24

3

14

3

24

3

11

21

4

3

24

1

121

66684

3),(

22284

1),(

8),(

2

1

x

x

x

xxxxx

x

xxfMPP

x

x

x

xxxxx

x

xxfMPP

xxxxfy

x

x

37

Note on MPP for Multiple Inputs When the MPP for a particular

input is zero, you have found a relative extrema point for the production function.

In general, the MPP w.r.t. input 1 does not have to equal MPP w.r.t. input 2.

38

The Isoquant An isoquant is the set of inputs that

give you the same level of output. To find the isoquant, you need to

set the dependent variable y equal to some number and examine all the combinations of inputs that give you that level of output. An isoquant map shows you all the

isoquants for a given set of inputs.

39

Example of An Isoquant Map: y = -x1

2+24x1-x2

2+26x2

0 4 8

12

16

20

24 0

3

6

9

12

15

18

21

24

300.00-350.00

250.00-300.00

200.00-250.00

150.00-200.00

100.00-150.00

50.00-100.00

0.00-50.00

-50.00-0.00

40

Example 2 of An Isoquant Map: y = 8x1

1/4 x23/4

0 2 4 6 8 10

0

1

2

3

4

5

6

7

8

9

10

60-80

40-60

20-40

0-20

41

Finding the Set of Inputs for a General Output Given y = -x1

2+24x1-x2

2+26x2

Suppose y = -x12+24x1-x2

2+26x2

We can solve the above equation for x2 in terms of y and x1.

yxx

yxx

yxx

a

acbb

yxx

yxxxx

xxxxy

1212

121

2

121

2

2

2

2

121

1212

22

2221

21

2416913x

2

496467626x

)1(2

)24)(1(4)26(26x

2

4x

24c and 26,b 1,a define weIf

02426

2624

42

Question From the previous example, does it

make economic sense to have both the positive and negative sign in front of the radical? No, only one makes economic sense; but

which one. You should expect that you will have an

inverse relationship between x1 and x2. This implies that for this particular function, you

would prefer to use the negative sign.

43

Finding the Set of Inputs for a General Output Given y = 8x1

1/4

x23/4

Suppose y = 8x11/4 x2

3/4

We can solve the above equation for x2 in terms of y and x1.

3/11

3/4

2

4

1

1

4

3

2

4

3

24

1

121

16

8

8),(

x

yx

x

yx

xxxxfy

44

Marginal Rate of Technical Substitution (MRTS) The Marginal Rate of Technical

Substitution tells you the trade-off of one input for another that will leave you with the same level of output. In essence, it is the slope of the

isoquant.

45

Finding the MRTS There are two methods you can

find MRTS. The first method is to derive the

isoquant from the production function and then calculate the slope of the isoquant.

The second method is to derive the MPP for each input and then take the negative of the ratio of these MPP.

46

Equivalency Between Slope of the Isoquant and the Ratio of MPP’s

MRTSdx

dx

dxdydxdy

dx

dy

dx

dx

i

1

2

x

x

2

1

x

x

x

1

2

2

1

2

1

i

MPP

MPP

equal is MPPeach for y in change heisoquant t on the are weSince

MPP

MPP

MPP that know We

isoquant theof Slope MRTS

47

Finding the MRTS Using the ratio of the MPP’s Given y = -x1

2+24x1-x22+26x2

Suppose y = -x12+24x1-x2

2+26x2

262

242

262

242

2624

2

1

22

11

2221

21

2

1

2

1

x

x

MPP

MPPMRTS

xx

yMPP

xx

yMPP

xxxxy

x

x

x

x

48

Finding the MRTS Using the Slope of the Isoquant Given y = -x1

2+24x1-x2

2+26x2

Suppose y = -x12+24x1-x2

2+26x2

2

1

121

1

1

2

12

1

121

1

2

2

1

121

11

2

2

1

1212

121

2

24169

12

x

x

242241692

1

x

x

2416913xx

x

2416913x

2

496467626x

asfunction w above for theisoquant that thefound wepreviously From

yxx

x

xyxx

yxx

yxx

yxx

49

Finding the MRTS Using the Slope of the Isoquant Given y = -x1

2+24x1-x22+26x2

Cont.

13

12

x

x

13

12

x

x

26169

12

x

x

262424169

12

x

x

262424169

12

x

x

2624y know we

24169

12

x

x

2

1

1

2

2

12

2

1

1

2

2

1

222

1

1

2

2

1

2221

211

21

1

1

2

2

1

2221

211

21

1

1

2

2221

21

2

1

121

1

1

2

x

x

x

x

xx

x

xxxxxx

x

xxxxxx

x

xxxx

yxx

x

50

Finding the MRTS Using the ratio of the MPP’s Given y = Kx1

x2

Suppose y = Kx1x2

1

212

11

12

11

121

21

1

121

2

21

11

21

2

1

2

1

x

xxxxxMRTS

xKx

xKx

MPP

MPPMRTS

xKxx

yMPP

xKxx

yMPP

xKxy

x

x

x

x

51

Finding the MRTS Using the Slope of the Isoquant Given y = Kx1

x2

Suppose y = Kx1x2

1

1

1

2

1

1

1

1

2

1

1

11

2

1

11

12

12

21

x

x

x

x

xx

x

xK

y

xK

y

xK

y

xK

y

Kx

yx

Kx

yx

xKxy

52

Finding the MRTS Using the Slope of the Isoquant Given y = Kx1

x2 Cont.

1

2112

1

2

121

2

1211

2

1

1

21

1

2

1

1

21

1

2

21

x

x

x

x

x

x

x

x

x

x

that know But we

x

xxx

xx

xxx

xK

xKx

xK

xKx

xKxy

53

Returns to Scale Returns to Scale examines what

happens to output when you change all inputs by the same proportion, i.e., f(tx1,tx2). There are three types of Returns to

Scale: Increasing Constant Decreasing

54

Increasing Returns to Scale Increasing Returns to Scale are said to

exist when f(tx1,tx2)>tf(x1,x2) for t > 1. This implies that as output is increasing,

the isoquants are getting closer together. Suppose y = f(x1,x2) = x1x2

This implies that f(tx1,tx2) = tx1tx2 =t2x1x2

Comparing f(tx1,tx2) and tf(x1,x2) implies f(tx1,tx2) = t2x1x2 >t f(x1,x2) = tx1x2, because when t >1, t2 > t.

55

Example Increasing Returns to Scale: y = 10x1

x2

0 2 4 6 8

100

1

2

3

4

5

6

7

8

9

10

900-1000800-900700-800600-700500-600400-500300-400200-300100-2000-100

56

Constant Returns to Scale Constant Returns to Scale are said to

exist when f(tx1,tx2)=tf(x1,x2) for t > 1. This implies that as output is increasing, the

isoquants are the same distance apart. Suppose y = f(x1,x2) = x1

½ x2½

This implies that f(tx1,tx2) = (tx1)½ (tx2)½ =tx1x2

Comparing f(tx1,tx2) and tf(x1,x2) implies f(tx1,tx2) = tx1x2 = t f(x1,x2) = tx1x2, because when t >1, t = t.

57

Example Constant Returns to Scale: y = 10x1

½ x2½

0 2 4 6 8

100

1

2

3

4

5

6

7

8

9

10 95-10090-9585-9080-8575-8070-7565-7060-6555-6050-5545-5040-4535-4030-35

58

Return to Scale Cont. Decreasing Returns to Scale are said to

exist when f(tx1,tx2)<tf(x1,x2) for t > 1. This implies that as output is increasing, the

isoquants are getting farther apart. Suppose y = f(x1,x2) = x1

¼ x2¼

This implies that f(tx1,tx2) = (tx1)¼ (tx2)¼

=t½x1¼x2

¼

Comparing f(tx1,tx2) and tf(x1,x2) implies f(tx1,tx2) = t½x1x2 < t f(x1,x2) = tx1

¼x2¼,

because when t >1, t½ < t.

59

Example Decreasing Returns to Scale: y = 10x1 ¼ x2

¼

0 2 4 6 8

100

1

2

3

4

5

6

7

8

9

10

28-3224-2820-2416-2012-168-124-80-4

60

The Multiple Product Firm Many producers have a tendency

to produce more than one product. This allows them to minimize risk by

diversifying their production. Personal choice.

The question arises: What type of trade-off exists for enterprises that use the same inputs?

61

Two Major Types of Multiple Production Multiple products coming from one

production function. E.g., wool and lamb chops

Mathematically: Y1, Y2, …, Yn = f(x1, x2, …, xn) Where Yi is output of good i Where xi is input i

62

Two Major Types of Multiple Production Cont. Multiple products coming from

multiple production functions where the production functions are competing for the same inputs. E.g., corn and soybeans

63

Two Major Types of Multiple Production Cont. Mathematically:

Y1= f1(x11, x12, …, x1m) Y2= f2(x21, x22, …, x2m) Yn= fn(xn1, xn2, …, xnm) Where Yi is output of good i Where xij is input j allocated to output Yi

Where Xj x1j + x2j + … + xnj and is the maximum amount of input j available.

64

Production Possibility Frontier A production possibility frontier (PPF)

tells you the maximum amount of each product that can be produced given a fixed level of inputs. The emphasis of the production

possibility function is on the fixed level of inputs.

These fixed inputs could be labor, capital, land, etc.

65

PPF Cont. All points along the edge of the

production possibility frontier are the most efficient use of resources that can be achieved given its resource constraints.

Anything inside the PPF is achievable but is not fully utilizing all the resources, while everything outside is not feasible.

66

Deriving the PPF Mathematically To derive the production possibility

frontier, you want to use the resource constraint on the inputs as a way of solving for one output as a function of the other.

67

PPF Example Suppose you produce two goods,

corn (Y1) and soybeans (Y2). Also suppose your limiting factor is

land (X1) at 100 acres. For corn you know that you have the

following production relationship: Y1 = x1

½

68

PPF Example Cont. For corn you know that you have

the following production relationship: Y2 = x2

½

We know that 100 = x1 + x2.

69

Solving PPF Example Mathematically

2

12

12

21

22

22

21

21

222

2

1

22

211

2

1

11

100

100

100

100

x Y

x

:following theknow We

YY

YY

YY

xx

Yx

YxY

70

PPF Graphical Example PPF

0

2

4

6

8

10

12

0 2 4 6 8 10 12

PPF

71

Marginal Rate of Product Transformation (MRPT) MRPT can be defined as the

amount of one product you must give up to get another product. This is equivalent to saying that the

MRPT is equal to the slope of the production possibility frontier.

MRPT = dY2/dY1

Also known as Marginal Rate of Product Substitution.

72

Find MRPT of the Following PPF: Y2 = (100-Y1

2)½

Suppose Y2 = (100-Y12)½

0100

2*1002

1

100

2

12

111

2

12

12

11

2

2

12

12

YYdY

dYMRPT

YYdY

dYMRPT

YY