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With best wishes of luck to our students.
2
Mathematical Economics
Third level
Economics Department
Prof. Dr. Ahmed Mandour
Professor of Economics
Faculty of Economic Studies and Political Science
Alexandria University
2023 / 2024
3
Table of Contents
Preface ....................................................................................................................................... 7
Chapter 1 ................................................................................................................................... 8
1)
Nature of Mathematical Economics.................................................................................. 8
A)
Definition................................................................................................................. 8
B)
Why use Mathematics?............................................................................................. 8
C)
Advantages of using Mathematical approach............................................................. 8
D)
Economic model....................................................................................................... 9
2)
Functions and Graphs .................................................................................................... 10
A)
Function ................................................................................................................. 10
B)
Definition of Domain and Range: ............................................................................... 11
C)
Inverse function: ........................................................................................................ 11
D)
Explicit and Implicit functions ................................................................................ 12
E)
Types of Functions: .................................................................................................... 13
F)
Graphing a Linear function:........................................................................................ 16
Chapter 2 ................................................................................................................................. 18
1)Demand Function: .......................................................................................................... 18
2)Supply Function: ............................................................................................................ 19
3)Proportional Income Tax Function .................................................................................. 20
4)Saving Function:............................................................................................................. 21
5)The budget surplus or deficit function: ........................................................................... 22
Chapter 3 ................................................................................................................................. 25
Linear economic model ......................................................................................................... 25
A)
A simple breakeven model ................................................................................... 25
B)
Partial market equilibrium linear model (linear model): staticanalysis ......................... 28
C)Simple National income equilibrium model (linear model) ............................................. 33
Chapter 4 .................................................................................................................................. 44
1)Non-Linear functions ....................................................................................................... 44
A)Economic analysis ......................................................................................................... 44
B)Types of functions: ........................................................................................................ 44
C)The Quadratic function: ................................................................................................. 45
D) Cubic Function: ............................................................................................................ 50
2)Rational functions .............................................................................................................. 51
A)Rectangular Hyperbola: ..................................................................................................... 51
Chapter 5 .................................................................................................................................. 55
Comparative analysis and the concept of derivative. .............................................................. 55
4
A)Measuring the point elasticity: ....................................................................................... 62
B)
Non-linear demand function: demand curves with constantprice elasticity............... 63
C)
Other types of demand elasticity: ............................................................................ 64
Chapter 6 .................................................................................................................................. 66
Using derivatives in economic analysis .................................................................................. 66
1)
Analyzing revenue: ...................................................................................................... 66
2)Relationship between MC and ATC:................................................................................... 67
3)Analyzing Production: ........................................................................................................ 68
4)Relationship between marginal product (MPL)and marginal cost (MC): ........................... 69
5) Income tax rate and Income Multiplier:.............................................................................. 70
6)Applying the concept of derivative on consumptionand saving functions: ............................. 70
Chapter 7 .................................................................................................................................. 72
Partial derivatives and application to comparative static analysis ........................................... 72
1)The concept of Partial derivatives ....................................................................................... 72
2)Techniques of partial differentiation ................................................................................... 73
3)Economic examples on partial derivatives .......................................................................... 74
4)Market Model .................................................................................................................... 75
5)National Income Model ...................................................................................................... 76
Chapter 8 .................................................................................................................................. 79
Optimization Relative Maximum and Minimum .................................................................... 79
1)
Definition of Relative maximum and minimum .............................................................. 79
2)Conditions of Relative maximum and minimum ............................................................. 80
3)Using Optimization in Economic analysis........................................................................... 81
Chapter 9 .................................................................................................................................. 89
Total differentials and total derivatives .................................................................................. 89
1)Total differential ................................................................................................................ 89
2)Total derivative .................................................................................................................. 90
3)Some economic examples on total differentials ................................................................... 91
Homogeneous functions and its economic applications .......................................................... 95
1)
Definition and properties (general) ................................................................................. 95
2)
Properties of linear homogeneity .................................................................................... 96
3)
Homogeneity of Cobb-Douglas function ........................................................................ 96
4) Relation between average and marginal productof inputs in Cobb-Douglass production
function ................................................................................................................................. 97
5)
Euler’s theorem and distribution of output...................................................................... 98
6)
Production partial elasticities and relative shareof labor and capital ................................ 98
7)
Expansion path for linear homogenous Cobb-Douglas.................................................... 99
5
Chapter 11............................................................................................................................... 102
Constrained optimization and lagrange multiplier model ...................................................... 102
1)
free and constrained optimization................................................................................. 102
2)
Finding the Optimum values assuming equalityconstraint............................................. 102
3)
Lagrange Multiplier Method ........................................................................................ 104
4)
Some Economic applications ....................................................................................... 106
a)
Utility Maximization. ............................................................................................... 106
b)Constrained output maximization:................................................................................. 107
c)Cost minimization: ....................................................................................................... 109
6
Preface
The use of mathematics in the study of modern Economics has
become essential at both undergraduate and postgraduate levels.
Mathematical Economics is especially important to understand economic
theories and problems, enabling the students how to construct a model,
understand applications, predicting what might happen in real world and
evaluate the effects of a given policies at micro or macro level.
This book covers the relevant tools and techniques of math that are
usually used in studying the quantitative relations between any economic
variables
I hope to enjoy and get benefit from studying this book at the
beginning level of mathematical economics.
I am thankful to my teaching assistant Alaa Elshazly for her endless
and sincere efforts in preparing.
Author
Professor Dr: Ahmed Mandour
7
Chapter 1
1) Nature of Mathematical Economics
A) Definition
Economics is a social science concerned with the relation between differentparts of the Economy,
many of these relations are quantitative in nature (relation between demand of the product and its
price, relation between consumption andincome).
Mathematical Economics study this quantitative relation to explain how the economy operates
and attempt to make predictions about what may happen to a given economic variables if certain
changes take place (example: What effect of a crop failure will have on crop prices, what effect
a given increase in sales tax will have on the price of goods. Mathematical Economics also gives
some guidelines that firms, governments, and individuals might follow if theywished to allocate
their scarce resources efficiently. Mathematical Economics makes use of mathematical tools to
derive a set of conclusions or theories from agiven set of assumptions.
B) Why use Mathematics?
•
To make quantitative predictions.
•
To make simplifications, to make certain concepts much easier and clearerto understand.
•
To obtain answers to optimization problem.
•
To estimate parameters of the models from statistical data (this branch iseconometrics).
C) Advantages of using Mathematical approach
•
The language used is more concise and precise.
•
Forcing us to state explicit assumptions (if – then), (condition – result).
•
Allow us to treat the general (n-variable) case.
•
Allow us to construct a simple model to understand very complicated problems in the real
world.
8
D) Economic model
Is a simplified presentation of the complex real world, by focusing on the basic features of a
situation under consideration. The main purpose of economic model is to derive a set of
conclusions which logically follow from a given assumptions.
1. Ingredients of a mathematical model:
It consists of a set of equations to describe the structure of the model.
The equation relates number of variables to one another in certain ways, theequation gives the
mathematical form to the set of analytical assumptions adopted.
• Variables, constants and parameters:
Variable: is something that can take any value, ex: 𝑷, 𝐓𝐂, 𝐓𝐑, 𝛑.
•Variables could be endogenous, when their solution values are determined from the model.
•Variables could be exogenous, when its value is determined by external forces outside of the
model (they are given) …ex: price is endogenous variable in market equilibrium model, price is
exogenous in other models as model of consumer choice.
•The equation of a consumer budget line:
𝑰 = 𝑷𝑿𝑿 + 𝑷𝒀𝒀
(𝑷𝑿 𝒂𝒏𝒅 𝑷𝒀) are given in this model.
Constant: has a given value and does not change 𝒚 = 𝒂 + 𝒃𝒙: 𝒂 is constant
When a constant is joined to a variable it called a coefficient like 𝒃 in the linear equation. (b) is
the slope it called a parameter.
2. Types of equations:
i.
Behavioral Equation: It specifies the manner in which a variable behave.
in response to changes in other variables……. ex:
𝑻𝑪 = 𝟕𝟓 + 𝟏𝟎𝑸
(Linear equation)
9
𝑻𝑪 = 𝟕𝟓 + 𝟏𝟎𝑸𝟐
(Quadratic equation)
The behavioral equation gives the mathematical expression of adoptedassumptions in the model.
ii. Equilibrium Equation: Exist only when the model involves the conceptof equilibrium (Qd
=Qs)
2) Functions and Graphs
A) Function
Is a mathematical relation between the values of two or more variables such that a unique value
of dependent variable exists for each value of independent variable orvariables. The function is a
single valued…. ex: 𝑦 = (𝑥) = 2𝑥
Sometimes we may have the same value of 𝑦 for more than value of 𝑥 …. ex: 𝑦 = 𝑦0 when 𝑥 =
𝑥1 𝑜𝑟 𝑥 = 𝑥2
When the 𝑥 value is given, it may not always be possible to determine a unique- value of( y. ex: 𝑦
≥ 2𝑥 (inequality), We call this a relation: many points on the line or in the shaded area satisfy the
relation.
10
We do not have a unique value of (y) for each value of (𝑥).
B) Definition of Domain and Range:
• The domain of the function y = F(x) is the set of all possible values thatindependent
variable (𝑥) can take.
• The range of the function is the set of all values that dependent variable (y) can take.
❖ Example: Suppose a specific demand function is, 𝑸𝒅
= 𝟏𝟐𝟎 − 𝟐𝑷
𝑄𝑑: Quantity Demanded.
𝑃: Price.
In Economics, we can restrict the Price and Quantity demanded to be non-negative.
Domain: (0 ≤ 𝑃 ≤ 60), Range: (0 ≤ 𝑄𝑑 ≤120)
C) Inverse function:
•Reverses the relationship in a function.
If the original function is 𝑦 = (𝑥), the inverse function is: 𝑥 = (𝑦)
(G) is a different function of (𝐹).
❖ Example: 𝑄𝑑 = 120 − 2𝑃
2𝑃 = 120 –𝑄𝑑
𝑷 = 𝟔𝟎 − 𝟎. 𝟓𝑸𝒅
(Inverse Function).
•Not all functions have inverse function, The mathematical necessary condition for afunction to
have inverse is that the original functionmust be “monotonic” this means as (x) increases (𝑦)
11
must either increases or decreases it cannot increase then decrease or vice-versa ..this will insure
that one unique value of (𝑦) exist for any value of (𝑥) also one unique value of (𝑥) for any value
of (𝑦).
❖ Example: 𝑦 = 9𝑥 − 𝑥2, 0 ≤ 𝑥 ≤ 9
Each value of (x) will determine a unique value of (𝑦), but some values of (𝑦) will have two
values of (𝑥). this function is not monotonic; (y) increases then decreases as (𝑥) increases.
𝑥
1
2
3
4
5
6
7
𝑦
8
14
18
20
20
18
14
When 𝒚 = 18, 𝒙 = 3 and 𝒙 = 6
D) Explicit and Implicit functions
If y = F(x) and x = G(y), It is possible to define which variable is dependent andindependent then
the function is Explicit.
❖ Example 𝑐 = 𝑎 + 𝑏,
OR 𝑏𝑦 = 𝑐 – 𝑎
[ c depends on 𝑦]
𝑎
𝑐
𝑏
𝑏
𝑦= +
[ y depends on c]
12
Note: Any explicit function can be transformed to implicit by transforming into
zero form.
𝑦=
𝑐 𝑎
+ = 0 (Implicit)
𝑏 𝑏
•Not all implicit can be written as explicit. ex: If 𝑦2 + 𝑥2 − 1 = 0 (implicit) the function is not a
single unique value because for the same value of (𝑥) → (𝑦) cantake more than one value.
If x = 0y = ±1
If x = 1y = 0 }
If x = −1y = 0
This is the equation of the circle.
E) Types of Functions:
•Algebraic functions consist of Polynomial Functions and Rational Functions.
1.polynomial functions
Refers to "multiterm" the general form of polynomial function of one variable (x) has the form:
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 + 𝑑𝑥3 … … … … … … 𝑛𝑥𝑛 , each term contains a coefficient as well as a nonnegative integer power of the variable (𝑥), thesecoefficients are called parameters the degree of
the function depends on the value ofthe power (𝑛).
Case 1: 𝑛 = 0
,
𝑦 = 𝑎𝑥0 = 𝑎
(Constant function)
13
Case 2: 𝑛 = 1
,
𝑦 = 𝑎 + 𝑏𝑥1
(Linear function)
Case 3: 𝑛 = 2
,
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
(Quadratic function)
Case 4: 𝑛 = 3
,
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 + 𝑑𝑥3
(Cubic function)
These types of the functions can be graphed as follow:
Case 1 (constant function)
Case 2 (linear function)
• The value of (𝑦) does not change as (𝑥)
• 𝑎: is 𝑦 intercept , 𝑎 > 0
changes this function could be AR function
• 𝑏: is the slope
under perfect competition firm, the price or AR
• 𝑏 > 0, The linear function is
is constant or fixed at any quantity of sales.
a straight-line upwardsloping.
• If b < 0 , its downward sloping straight line.
14
Case 3 (quadratic function)
Case 4 (cubic function)
• 𝑎 > 0, 𝑏 > 0, 𝑐 < 0
• 𝑎 > 0, 𝑏 > 0, 𝑐 < 0, 𝑑 >0
• If 𝑐 > 0, the curve opens up.
1. Rational Functions
Is represented by the ratio of two polynomial functions in the same variable (𝒙)
❖ Example
𝒂 + 𝒃𝒙
→ Is rational function →a, b, c, d are parameters.
𝒂 + 𝒄𝒙 + 𝒅𝒙𝟐
Rectangular hyperbola has interesting applications in Economics:
→ (a) is the constant function, (x) is the linear equation.
𝑦=
𝑎
Constant function
→𝐎𝐑 y. x = a = constant (
), (when a = 0, b = 1)
𝑥
Linear function
Note:
•This function could be used as → (AFC * Q) = FIXED COST
OR
•Unit elastic demand curve when demand is non-linear.
𝑃. 𝑄 = 𝑎 (Constant)
15
(𝑎) The consumer expenditure is constant
F) Graphing a Linear function:
The linear relation between two variables (x) and (y) is represented by the function:
𝑦 = 𝐹(𝑥) = 𝑎 + 𝑏𝑥
where a and b are parameters.
Assuming 𝑎 and 𝑏 are positive values then we obtain the graph of the function as a straight line.
Set 𝑏 = 0→
𝑦=𝑎
Set 𝑥 = 0 →
0=𝑎+𝑏𝑥
→ − a = bxx = −
𝑎
𝑏
16
•We call (a) to be (𝑦) intercept.
𝑎
• (− ) is (x) intercept.
𝑏
• (𝑏) is the slope of the straight line which is constant
At point (A): 𝑦1 = 𝑎 + 𝑏𝑥1
At point (B): 𝑦2 = 𝑎
𝑏=
∆y
>0
∆x
𝑦2 − 𝑦1 = 𝑎 + 𝑏𝑥2 − 𝑎 − 𝑏𝑥1
= 𝑏𝑥2 − 𝑏𝑥1
∆𝑦 = 𝑏 (𝑥2 − 𝑥1)
𝑏=
= (∆𝑥)
∆y BC
=
∆x AC
The slope remains constant throughout the line.
17
Chapter 2
Using linear function in economic analysis:
(Some examples from micro and macroeconomics)
1)Demand Function:
Linear demand function can be written as: 𝑸𝒅 = 𝒂 − 𝒃𝑷
•𝑄𝑑 is the Quantity demanded for a given product.
• (𝑎) and (𝑏) are parameters and constants since price (𝑃) is the independent
variable, (𝑃) should be taken as x-axis and (𝑄𝑑) is dependent variable should
be taken as y-axis.
Graphing the linear demand function:
Set (𝑃) equal zero in the equation (𝑎) isy-intercept
(𝑎 > 0).
Set (𝑄𝑑) equal zero in the equation: 0 =𝑎 – 𝑏𝑃
𝑎 = 𝑏𝑝→ 𝑃 =
𝑎
𝑏
𝑎
• ( )is x-intercept.
𝑏
𝑎
Connecting the two points (, ) we get straight-line
𝑏
demand curve.
•The slope is given by (−b), the slope is negative and constant.
•Demand is downward sloping because the inverse relation between (𝑃) and
(𝑄𝑑) this function assume, other factors affecting (𝑄𝑑) heldconstant
18
(𝐼, 𝑃𝑋, 𝑃𝑌 … …).
•Change in (P) causes movement along the demand curve.
2)Supply Function:
Linear supply function can be written as: 𝑸𝒔 = −𝒄 + 𝒅𝑷
•Qs: Is the Quantity supplied of given product.
•P: Is the Price of this product.
• (𝑐, 𝑑 > 0)
• (c, d) are parameters and constant.
• (P) should be taken as x-axis, it is the independent variable.
• (𝑄𝑠) should be taken as y-axis it is the dependent variable.
•Graphing the linear supply function:
•Set (𝑃) equal zero in the equation, we get 𝑄𝑠 = −𝑐 (y-intercept)
Graphing the linear supply function:
•Set (𝑄𝑠) equal zero, we get 0 = −𝑐 + dP, dP =
c ,P =
𝑐
𝑑
•Connecting the 2 points (−𝑐,
𝑐
𝑑
) and make
extension we get the upwardssloping straight line.
•( 𝑐) represents the minimum Price, the firm will
charge below this 𝑑 price no production will cover
𝑐
its costs (𝑃 ≥ ).
𝑑
•The slope is positive and constant and equal (𝑑) the linear supply function
19
assumes other factors affect 𝑄𝑠 except price are held constant (input prices,
technology, taxes...), Change in (𝑃) causes movement,change in other factors
causes shift of supply curve.
3)Proportional Income Tax Function
•Suppose the tax collection (T) is proportional
to Income (Y)
𝑇 = 𝑡. 𝑌
(0 < 𝑡 < 1)
•𝑡: tax rate
Suppose (𝑡 = 0.3), then 𝑇 = 0.3 𝑌
The function is linear, the slope = 0.3.
•The effect of income tax on disposable income:
𝑌𝐷 is the disposable income “It is the income after
payingtaxes”.
𝑌𝐷 = 𝑌 − 𝑇 = 𝑌 − 𝑡𝑌 = 𝑌 (1 − 𝑡)
𝑌𝐷 = 𝑌 (1 − 0.3) = 0.7 𝑌
•The effect of income tax on consumption:
Suppose: 𝐶 = 𝑎 + 𝑏𝑌𝐷
If 𝑎 = 100, 𝑏 = 𝑀𝑃𝐶 = 0.8
Then 𝐶 = 100 + 0.8𝑌𝐷
𝐶 = 100 + 0.8 (𝑌 − 𝑇)
𝐶 = 100 + 0.8 (𝑌 − 𝑡𝑌)
20
𝐶 = 100 + 0.8 (1 − 𝑡) 𝑌
𝐶 = 100 + 0.8 (1 − 0.3) 𝑌
𝐶 = 100 + 0.8 (0.7) 𝑌
𝐶 = 100 + 0.56 𝑌
4)Saving Function:
•Saving (𝑆) can be defined as: the part of disposable income (𝑌𝐷) that is not
spent.
𝑆 = 𝑌𝐷 − 𝐶
•Suppose:
𝐶 = 𝑎 + 𝑏𝑌𝐷
𝑆 = 𝑌𝐷 − 𝑎 − 𝑏𝑌𝐷
= −𝑎 + (1 − 𝑏) 𝑌𝐷
(0 < 𝑏 < 1)
•The slope is (1 − 𝑏) > 0 (1 − 𝑏) = 𝑀𝑃𝑆
•If 𝑌𝐷 is zero, 𝑆 = −𝑎, 𝐶 > 𝑌𝐷
•If 𝑆 is zero, 0 = −𝑎 + (1 − 𝑏) 𝑌𝐷
𝑎 = (1 − 𝑏) 𝑌𝐷
𝑌𝐷 =
𝑎
𝑎
=
1 − 𝑏 𝑀𝑃𝑆
21
•When (𝑆) is zero the level of disposable income
𝑎
1−𝑏
is called the breakeven
income.
𝐼𝑓 𝑌𝐷 >
𝑎
…S > 0
1−𝑏
𝐼𝑓 𝑌𝐷 >
𝑎
…S > 0
1−𝑏
Note: In this example the economic variable could be negative, and negative
quantity make sense, so not all economic variables are required to be positive.
5)The budget surplus or deficit function:
The Budget surplus (Bus) is defined as the excess of government revenues
consisting of taxes (T) over its total expenditures (𝐺), A negative Bus is budget
deficit.
𝐵𝑢𝑠 = 𝑇 − 𝐺
We want to study Bus in relation to the level of income 𝑌
Assuming a proportional income tax that gives a tax revenue 𝑇 = 𝑡. 𝑌
𝑡: Is the income tax rate (1 > 𝑡 > 0)
𝑇T: The total tax collection.
𝐵𝑢𝑠 = 𝑡𝑌 − 𝐺̅
As 𝐺 = 𝐺̅
Set 𝑌 = 0 …. 𝐵𝑢𝑠 = − 𝐺̅ (Budget Deficit) Set 𝐵𝑢𝑠 = 0 …. 𝑡𝑌 = 𝐺̅ (Balanced
Budget)
•When (𝑌) exceeds 𝑌∗ Budget surplus exists.When (𝑌) is below 𝑌∗ Budget
22
deficit exist.
The effects of Government purchases on the (𝐵𝑢𝑠):
•An increase in (G) must reduce the (𝐵𝑢𝑠) or increase Budget deficit as (𝑇) is
constant in the equation.
𝐵𝑢𝑆 = 𝑇 − 𝐺 𝖳
However, an increase in (𝐺) will give rise to a multiple increase in income (𝑌)
due to expenditure multiplier effect and (𝑡𝑌) will increase, As income
increases Tax collection will increase and (𝐵𝑢𝑠) may increase, This can be
explained as follows:
𝐵𝑢𝑆 = 𝑇 − 𝐺
𝑇 = 𝑡. 𝑌
∆𝐵𝑢𝑆 = ∆𝑇 − ∆𝐺
∆𝑇 = 𝑡. ∆𝑌
∆𝐵𝑢𝑆 = 𝑡. ∆𝑌 − ∆𝐺
∴ ∆𝑌 = 𝐾. ∆𝐺
•k is the government spending multiplier
∆𝐵𝑆 = 𝑡. 𝐾. ∆𝐺 − ∆𝐺
∆𝐵𝑆 = ∆𝐺[𝑡. 𝐾 − 1]
• (1 > 𝑏, 𝑀𝑃𝐶 > 0)
• (1 > 𝑡 > 0)
∆BuS = ∆G[t.
1
1−𝑏(1−𝑡)
− 1]
23
𝐾=
1
1−𝑏(1−𝑡)
∆BuS =
∆BuS =
∆G [t − [1 − b (1 − t)]
1−𝑏(1−𝑡)
∆G [t−1+ b− bt]
∆BuS =
1−𝑏(1−𝑡)
=
=
∆G [t − (1 − b +bt)
1−𝑏(1−𝑡)
−∆G [−t + 1 − b +bt]
1−𝑏(1−𝑡)
−∆G [(1 − t) − b (1 − t)] −∆G (1 − t) (1 − b)
=
<0
1 − 𝑏(1 − 𝑡)
1 − 𝑏(1 − 𝑡)
•As ∆𝐺 increase, ∆𝐵𝑢𝑆 ↓
(1 − t) (1 − b)
∆BuS
=−
−< 1
∆G
1 − 𝑏(1 − 𝑡)
Ex. Suppose G=100, b=0.8, t=0.2. what is the  Bus.
∆Bus=-100(1- 0.2) (1-0.8)/ 1-0.8(1-0.2)
∆Y = K. ∆G= 1 / 1-0.8(1-0.2).100= 278
t. ∆Y = 278 × 0.2 = 55.6
•An increase in G causes less increase in budget deficit.
24
Chapter 3
Linear economic model
We have already studied how linear functions can be used in economic
analysis. Inthis section we develop some economic models that depend on
linear function, we need to reach general solution (reduced form) that can be
applied to any situation regardless of the precise numerical parameters used
in each model.
We will develop three models:
A) A simple breakeven model.
B) Partial market equilibrium linear model.
C) National income equilibrium model.
A) A simple breakeven model
Since breakeven is where profit is zero, Profit (𝜋) is defined as the
difference between total revenue (𝑇𝑅) and total cost (𝑇𝐶).
We will assume the firm is selling its product in a perfectly competitive market
theproduct price is given at (𝑃) per unit.
𝑇𝑅 = 𝐹(𝑄) = 𝑎 + 𝑏𝑄 = 𝑃. 𝑄
•𝑎 = 0, 𝑏 = 𝑃
𝑇𝑅 Would go through the origin.
𝑇𝐶 = 𝐹(𝑄) = 𝑎 + 𝑏𝑄 = 𝑃. 𝑄
•𝑎 > 0, 𝑏 > 0
•𝑎 = Fixed cost
•𝑏 = Variable cost per unit of output
The breakeven point requires 𝑻𝑹 = 𝑻𝑪
25
𝑃. 𝑄 = 𝑎 + 𝑏𝑄
𝑄∗ =
𝑎
, This gives the quantity of output such that 𝑇𝐶 = 𝑇𝑅 and total profit
𝑝−𝑏
(𝜋 = 0).
𝑎 = 𝑃. 𝑄 − 𝑏. 𝑄 = 𝑄(𝑃 − 𝑏)
How 𝑄∗ will change when 𝐹𝐶, 𝐴𝑉𝐶, 𝑃 change?
𝑇𝑂 ∗
𝖳 Fixed cost
↓ Price− ↓ Unit variable cost
Graphically
Solution is represented by point of intersection of 𝑇𝑅 and 𝑇𝐶 lies at (𝑄∗) at
point (𝐵):
𝑇𝑅 = 𝑇𝐶, 𝜋 = 0
❖ Test yourself: Solve the numerical example we had before:
•𝜋 = 0
𝑂∗ =
100
100
=
= 40, TR = (40) (7.5) = 300, TC
1.5 − 5 2.5
= 100 + 5(40) = 300
When the firm think to reduce Fc to 90, how this will affect the breakeven?
26
The new breakeven point:
=
90
= 36 𝑢𝑛𝑖𝑡𝑠
2.5
𝑂∗ =
90
1.5 − 5
𝑇𝑅 = (7.5) ( 36) = 270
𝑇𝐶 = 100 + 5(36) = 270
•𝜋 = 0
Suppose the price now is 10$, Given the initial Tc function, what is the
breakevenpoint?
𝑂∗ =
100
100
=
= 20 𝑢𝑛𝑖𝑡𝑠
10 − 5
5
𝑇𝑅 = (20)(10) = 200
𝑇𝐶 = 100 + 5(20) = 200
•𝜋 = 0
As price increase the breakeven level of sales decreases as TC decrease.
27
❖ Test yourself: Solve the numerical example we had before:
•what is the effect of increase in price on break-even???
•What is the effect of increase in cost in breakeven???
B) Partial market equilibrium linear model (linear model): staticanalysis
The case of one good: (Partial)
We have three variables in the model:
𝑄𝑑, 𝑄𝑠, 𝑃
The equilibrium condition in the model
𝑄𝑑 = 𝑄𝑠
How 𝑸𝒅 and 𝑸𝒔 are determined?
We assume 𝑄𝑑 is a decreasing linear function of (𝑃) and 𝑄𝑠 is an increasing
linear function of (𝑃), provided no 𝑄𝑠 unless the price exceeds a particular
positive value.
The model consists of two behavioral equations which govern the demand and
supply side in the market.
The model can be written as:
𝑄𝑑 = 𝑎 − (𝑃)
(1)
𝑄𝑠 = −𝑐 + (𝑃)
(2)
𝑄𝑑 = 𝑄𝑠
(3)
•𝑎, 𝑏 > 0
•𝑐, 𝑑 > 0
All four parameters 𝑎, 𝑏, 𝑐, 𝑑 are positive.
The model can be presented graphically as follow:
28
The solution of the model at point (E) where 𝑄𝑑 = 𝑄𝑠
Equilibrium quantity 𝑄̅ = 𝑄𝑑 = 𝑄𝑠
Equilibrium price 𝑃`
Mathematical solution of the equation (reduced form):
𝑄𝑑 = 𝑄𝑠
𝑎 − 𝑏𝑃 = −𝑐 + 𝑑𝑃
𝑎 + 𝑐 = 𝑏𝑃 + 𝑑𝑃
𝑎 + 𝑐 = (𝑏 + 𝑑)𝑃
Since (𝑏 + 𝑑 ≠ 0), it is possible to divide both sides of the equation by (b+ d)
We get: P` =
𝑎+𝑐
𝑏+𝑑
(𝑃 > 0) Since 𝑎, 𝑏, 𝑐, 𝑑 > 0 by substituting in 𝑄𝑑 𝑜𝑟 𝑄𝑠
equations we get
= 𝑎 – 𝑏𝑃.
𝑄 = 𝑎−𝑏(
𝑄=
𝑎+𝑐
𝑎(𝑏 + 𝑑 ) − 𝑏(𝑎 + 𝑐)
)=
𝑏+𝑑
𝑏+𝑑
𝑎𝑏 + 𝑎𝑑 − 𝑎𝑏 − 𝑏𝑐
𝑎𝑑 − 𝑏𝑐
=(
)
(𝑏 + 𝑑)
𝑏+𝑑
Since (𝑏+ 𝑑 > 0), the positive value requires that 𝑎𝑑 > 𝑏𝑐 to be economically
meaningful (to make sense).
29
Point (E) is the only stable position: (𝑄𝑑 = 𝑄𝑠 ) when (𝑃=𝑃) because a higher
price than 𝑃 (to the right of 𝑃 express excess supply).
(𝑄𝑠 > 𝑄𝑑 ): down pressure on price towards 𝑃̅ any lower price (to the left of
𝑃̅ express excess demand).
𝑄𝑑 > 𝑄𝑠: upward pressure on price toward (𝑃̅ ).
❖ Test yourself: solve using the equilibrium reduced formula at the
example we had before:
When 𝑄𝑑 = 100 − 5𝑃,
𝑄𝑠 = −100 + 20𝑃
Make sure that the restrictions required in this model are satisfied.
The effects of an excise tax on competitive market equilibrium (a competitive
staticanalysis):
We could develop the previous model to consider the effects of imposing a
sales tax by government on the product we assume the amount of tax is
constant amount on each unit regardless of the price, it is called an excise tax
the producer must pay thegovernment the tax has been imposed. So, excise tax
will have effect only on supplyequation, the demand equation will not change.
30
Considering the effects of such taxes will bring the analysis to a form called
comparative static analysis, we will consider two static equilibrium positions,
beforeand after the tax, we shall be able to compare these to explain the excise
tax effectson market equilibrium.
Suppose the amount of the tax is (t) percent of sales and (t>0) and constant.
The firm will receive the market price (P) but must pay (t) to government, so
the price after paying tax is:
𝑃𝑡 = 𝑃 − 𝑡
Suppose the supply equation:
𝑄𝑠 = −𝑐 + 𝑑𝑃𝑡
Now will be,
𝑄𝑠 = −𝑐 + 𝑑(𝑃 − 𝑡)
𝑄𝑑 = 𝑎 − 𝑏𝑃 (𝑎 > 0, 𝑏 > 0)
𝑄𝑠 = −𝑐 + 𝑑 (𝑃 − 𝑡)
(𝑐, 𝑑 > 0, 𝑡 > 0)The equilibrium condition:
𝑄𝑑 = 𝑄𝑠
The equilibrium imposing tax:
𝑎 − 𝑏𝑃 = −𝑐 + 𝑑(𝑃 − 𝑡)
𝑎 − 𝑏𝑃 = −𝑐 + 𝑑𝑃 − 𝑑𝑡
𝑎 + 𝑐 + 𝑑𝑡 = 𝑑𝑃 + 𝑏𝑃 = 𝑃(𝑑 + 𝑏)
𝑃=
a + c + dt
𝑎+𝑐
𝑑𝑡
=(
)+(
)
d + b
𝑑+𝑏
𝑑+𝑏
Substituting in Qd or Qs, we get:
𝑄 = 𝑎 − 𝑏 [(
𝑎+𝑐
𝑑+𝑏
)+(
𝑑𝑡
𝑑+𝑏
)]
31
•1 >
dt
d+b
>0
𝑄=
𝑎(𝑑 + 𝑏) − 𝑏𝑎 − 𝑏𝑐 − 𝑏𝑑𝑡 𝑎𝑑 + 𝑎𝑏 − 𝑎𝑏 − 𝑏𝑑𝑡
=
𝑑 + 𝑏
𝑑 + 𝑏
𝑄=(
𝑎𝑑 − 𝑏𝑐
𝑏𝑑𝑡
)−(
)
𝑑+𝑏
𝑑+𝑏
Notice:
When you set t=0 in 𝑄, 𝑃̅ formula you get the previous formula before imposing
tax.
•Imposing tax will cause an increase in equilibrium price 𝑃 because
and decrease in equilibrium quantity by the amount (
𝑏𝑑𝑡
𝑑+𝑏
dt
d+b
).
•Equilibrium price will increase by less than the amount of the tax
1, 𝑎𝑠 𝑑 + 𝑏 𝑑.
Decrease in equilibrium quantity by (
𝑏𝑑
𝑑+𝑏
).t
❖ Test yourself: solve the previous example:
Assuming excise tax=2$ per unit
𝑄𝑑 = 100 − 5𝑃
𝑄𝑠 = −100 + 20𝑃
(before taxes)
Supply will be:
𝑄𝑠 = −100 + 20(𝑃 − 𝑡) 100 − 5𝑃 = −100 + 20(𝑃 − 2)
100 − 5𝑃 = −100 + 20𝑃 − 40
240 = 25𝑃
𝑃=
240
= 9.6
25
32
> 0,
d
d+b
>
The increase in equilibrium price (1.6) $
𝑄 = 100 − 5(9.6) = 52
Equilibrium quantity decreases from 60 to 52.
Using the reduced formula:
𝑃=(
𝑄=
100 + 100
20
240
)+(
)×2=
=9
20 + 5
20 + 5
25
(100)(20) − (5)(100) (5)(20)
130
×2=
= 52
(5 + 20)
(5 + 20)
25
The effects of excise tax graphically:
Tax-incidence:
Consumer 1.6$ (80% of the taxes seller), “1.6$ of 2$ =80%”
Q
P
C)Simple National income equilibrium model (linear model)
Assumptions
• Closed economy.
• Taxes are functions of income 𝑇 = 𝑡𝑌.
33
• No money sector.
• 0 < 𝑡 < 1 Income tax rate.
• 𝐼I and G are autonomous.
• 1 > 𝑀𝑃𝐶 > 0.
❖ Equilibrium income (Y)
𝑌=𝐶+𝐼+𝐺
𝐶 = 𝑎 + 𝑏𝑌𝑑
𝑌𝑑 = 𝑌 − 𝑇 = 𝑌 – 𝑡𝑌
Y = a⏟ + b(Y − tY) + I̅ + G̅
𝑐
❖ Solving for 𝒀:
𝑌 − 𝑏(𝑌 − 𝑡𝑌) = 𝑎 + 𝐼 + 𝐺
𝑌 − 𝑏𝑌 + 𝑏𝑡𝑌 = 𝑎 + 𝐼 + 𝐺
[1 − 𝑏 + 𝑏𝑡] = 𝑎 + 𝐼 + 𝐺
Y =
Y =
∆𝑌
∆𝐺
1
(𝑎 + 𝑇 + 𝐺)
1 − 𝑏 + 𝑏𝑡
1
(𝑎 + 𝑇 + 𝐺)
1 − 𝑏(1 − 𝑡)
=
∆𝑌
∆𝐼
=
1
=𝐾
1 − 𝑏(1 − 𝑡)
𝐾 is 𝐺, 𝐼 multiplier…. 0 < 𝐾 > 1, an increase in (𝐺) or (𝐼) by 1$ (holding 𝑎
constant) will increase the equilibrium level of income 𝑌 by
greater than (1).
Example:
34
1
1−𝑏(1−𝑡)
which is
If 𝑏 = 𝑀𝑃𝐶 = 0.75 and 𝑡 = 0.2, the 𝐺 multiplier is?
K =
1
1
=
= 2.5
1 − 0.75(1 − 0.2) 1 − 0.75(0.8)
∆𝑌= ∆𝐺. 𝐾 = (1) (2.5) = 2.5
35
Chapter 4
1)Non-Linear functions
A)Economic analysis
The relationship between variables used in economic analysis is not always
linear or straight-line functions, many of such relationships can be realistically
represented onlythrough the use of non-linear functions, we will study the
following non-linear functions:
1) Quadratic function.
2) Cubic function.
3) Rectangular hyperbola.
4) Functions of more than one independent variable (Cobb-Douglas
function).
In general, a function with multi term (polynomial functions) has the following
form….
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 + 𝑑𝑥3 … … … … … … 𝑛𝑥𝑛
B)Types of functions:
•Constant function: 𝑦 = 𝑎𝑥0 = 𝑎
•Linear function: 𝑦 = 𝑎 + 𝑏𝑥1
•Quadratic function: 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
•Cubic function: 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 + 𝑑𝑥3
The powers of (X) are called exponents, it determines the degree of the
function.
44
C)The Quadratic function:
Using the quadratic function in economic analysis:
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
𝑎, 𝑏 > 0, 𝑐 ≶ 0
𝑐>0
𝑐<0
𝑦 = 1 + 4𝑥 − 2𝑥2
𝑦 = 1 + 4𝑥 + 2𝑥2
X
3
2
1
0
-1
X
3
2
1
0
-1
-2
Y
-5
1
3
1
-5
Y
31
17
7
1
-1
1
45
I.
Break-Even analysis
❖ Example: Assume a firm faces this situation, such that:
𝑇𝑅 = (𝑄) = 260𝑄 (Linear)
TC
$
𝑇𝐶 = (𝑄) = 1125 = 10𝑄 + 5𝑄2 (Quadratic)
TR
11700
𝑇𝑅 = 𝑇𝐶
260𝑄 = 1125 + 10𝑄 + 5𝑄2
1125
Q
𝜋 = 5𝑄 − 260𝑄 + 10𝑄 + 1125 = 0
2
π
5
25
45
𝜋 = 5𝑄2 − 250𝑄 + 1125 = 0 (5𝑄 − 25) (𝑄 − 45) = 0
𝑄 = 5 𝑂𝑅 𝑄 = 45
2000
Q
5
25
45
The quadratic equation can be solved by using thisformula (the function has
two roots “two values ofX when Y=0)
−𝑏 ± √𝑏2 − 4𝑎𝑐
𝑋=
2𝑐
•𝑏2 > 4𝑎𝑐
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝜋 = 260𝑄 − 1125 − 10𝑄 − 5𝑄2 = 0
𝜋 = −1125 + 250𝑄 − 5𝑄2 = 0
46
π
•𝑎 = −1125, 𝑏 = 250, 𝑐 = −5
−250 ± √(250) 2 − 4(−1125)(−5)
𝑋=
2(−5)
𝑋=
𝑋=
−250 ± √62500 − 22500
−10
−250 ± √40000 −250 + 200 −50
=
=
=5
−10
−10
−10
OR
𝑋=
−250 ± √40000 −250 + 200 −450
=
=
= 45
−10
−10
−10
Notice:
The output that maximizes profits 𝑀𝑅 = 𝑀𝐶, will be at midway between the two
roots of quadratic profit function (between 5 & 45) it will be at 25?
𝑀𝑅 = 260, 𝑀𝐶 =
𝑑𝑇𝑐
𝑑𝑄
= 10 + 10𝑄
260 = 10 + 10𝑄
10𝑄 = 250
𝑄 = 25
𝜋 = 2000
𝑇𝑅 = 25(260) = 6500, 𝑇𝐶 = 4500
Proof the Quadratic formula:
−𝑏 ± √𝑏2 − 4𝑎𝑐
𝑋=
2𝑐
47
•𝑏2 > 4𝑎𝑐
Suppose: 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
Set 𝑦 = 0
0 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
𝑏𝑥 + 𝑐𝑥2 = −𝑎
Dividing by (𝑐)
𝑏𝑥
−𝑎
+ 𝑌2 =
𝑐
𝑐
Add and subtract (
𝑏2
4𝑐 2
)
𝑏𝑥 𝑏2
−𝑎 𝑏2
𝑥 +
+ 2=
+ 2
𝑐
4𝑐
𝑐
4𝑐
2
Can be written as:
𝑏 2
𝑏2 𝑎
(𝑥 + ) = 2 −
2𝑐
4𝑐
𝑐
Take the square root of both sides:
√𝑏2 ± 𝑎 √𝑏2 − 4𝑎𝑐
𝑏
𝑥+ 𝑐 = 𝑐
=
2
4 ±𝑐
4𝑐 2
𝑥=
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑐
−𝑏 ± √𝑏2 4𝑎𝑐
𝑥=
2𝑐
Finding the vertex:
𝑥=
−𝑏
2𝑐
48
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2
𝑑𝑦
= 𝑏 + 2𝑐𝑥 = 0(max 𝑜𝑟 min )
𝑑𝑥
2𝑐𝑥 = −𝑏
𝑥=
II.
−𝑏
2𝑐
Partial market equilibrium
𝑥 =−𝑏 2𝑐
Suppose the demand function is non-linear (quadratic) but supply function is
linear.
𝑄𝑑 = 4 − 𝑃2
𝑄𝑠 = 4𝑃 − 1
𝑄𝑑 = 𝑄𝑠
OR
𝑄𝑑 − 𝑄𝑠 = 0 4 − 𝑃2 − 4𝑃 + 1 = 0
−𝑃2 − 4𝑃 + 5 = 0
𝑃2 + 4𝑃 − 5 = 0
•𝑎 = −5, 𝑏 = 4, 𝑐 = 1
•The function opens up as 𝑐 > 0
−𝑏 ± √𝑏2 − 4𝑎𝑐
𝑃=
2𝑎
𝑃=
−4 ± √16 − 4(−5)(1)
2(1)
49
𝑃=
𝑃1 =
𝑃2 =
−4 ± √36
2
−4 + 6
=1
2
−4 − 6
= −5
2
𝑄𝑑 − 𝑄𝑠 = 0, when 𝑃 = 1 𝑜𝑟 𝑃 = −5
•Graphical solution
𝑄𝑑 = 4 − 𝑃2
𝑄𝑠 = 4𝑃 − 1
𝑄𝑑 = 𝑄𝑠
4𝑃 − 1 = 4 − 𝑃2
𝑃2 − 4𝑃 + 1 + 4 = 0
𝑃2 + 4𝑃 − 5 = 0 (𝑃 + 5) (𝑃 − 1) = 0
𝑃1 = 1, 𝑃2 = −5
When 𝑅1 = 1, 𝑄 = 4(1) − 1 = 3, OR 𝑄= 4 − 1 = 3
D) Cubic Function:
𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 + 𝑑𝑥3
𝑇𝐶 = 𝑎 + 𝑏𝑄 + 𝐶𝑄2 + 𝑑𝑄3
•𝑎, 𝑏, 𝑑 > 0
•𝑐 < 0
•3𝑑𝑏 > 𝑐2
50
𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶
𝐹𝐶 = 𝑎
𝑉𝐶 = 𝑏𝑄 + 𝑐𝑄2 + 𝑑𝑄3
𝑀𝐶 =
𝑑𝑇𝐶
= b + 2cQ + 3dQ2(Quadratic)
𝑑𝑄
Since 𝑇𝐶 is increasing as (𝑄) increases (𝑀𝐶) must take positive values for
non-negative (𝑄), so the minimum value of (𝑀𝐶) must be positive.
The minimum is given when
𝑑𝑀𝐶
𝑑𝑄
= 2𝑐 + 6𝑑𝑄 = 0 (Necessary condition)
2𝑐 = −6𝑑𝑄
𝑐 = −3𝑑𝑄
𝑑 2𝑀𝐶
= 6d > 0 (Sufficient condition)
𝑑𝑄 2
MC is minimum when Q =
−𝐶
3𝑑,
since 𝑄 > 0 and when 𝑑 > 0, 𝑐 must be
negative (𝑐 < 0).
𝑀𝐶 (𝑎𝑡 𝑚𝑖𝑛) = 𝑏 + 2𝑐 (
−𝑐
−𝑐
−𝑐 2
) + 3𝑑 ( ) 3𝑑 ( )
3𝑑
3𝑑
3𝑑
3𝑑𝑏 − 2𝑐 2 + 𝑐 2 3𝑑𝑏 − c2
𝑀𝐶 =
=
3𝑑
3𝑑
𝑐 < 0, 𝑑 > 0 is not sufficient to have 𝑀𝐶 > 0.
•The sufficient condition for positive minimum value of 𝑀𝐶 is (3𝑑𝑏 > 𝑐2).
2)Rational functions
A)Rectangular Hyperbola:
The function: 𝑦 =
𝑎
𝑋
•𝑎 is constant.
51
OR
𝑦. 𝑥 = 𝑎
OR 𝑦 = 𝑎𝑥−1
This function has many applications in economics: demand curve, with
price and quantity for which the total expenditure (p*Q) is constant.
𝑃. 𝑄 = 𝑎
❖ Example: Demand Function
P
Q
TR
20
5
100
10
10
100
5
20
100
4
25
100
2
50
100
Such curve is called unit elastic demand
Q
curve or rectangular hyperbola.
•Suppose total expenditure is constant at
(100), how to graph this demandcurve?
(𝑃. 𝑄) is constant along any point on the
demand curve, the curve is downward
sloping never meet the axes (any
rectangle below the curve has the same
area).
52
P
•
Another application: 𝐴𝐹𝐶 curve
𝐹𝐶 = 𝐴𝐹𝐶 × 𝑄 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The curve is downward sloping 𝐴𝐹𝐶 is decreasing as (𝑄) increases.
❖ Example
𝐹𝐶 = 1000
At 𝑄 = 100, 𝐴𝐹𝐶 = 10 At 𝑄 = 200, 𝐴𝐹𝐶 =
5 At 𝑄 = 500, 𝐴𝐹𝐶 = 2 Notice:
The area of any rectangular hyperbola
below the curve is constant =
𝐴𝐹𝐶 × 𝑄 = 𝐹𝐶
•
Another application: 𝐴𝑇𝐶 function
When 𝑇𝐶 function is linear
𝐴𝑇𝐶 =
𝑇𝐶
𝑄
Suppose: 𝑇𝐶 = 𝑎 + 𝑏𝑄
•𝑎 > 0 (constant)
•𝑏 > 0
𝐴𝑇𝐶 =
𝑎
+𝑏
𝑄
53
❖ Example
Suppose 𝑇𝐶 = 40 + 3𝑄
𝐴𝑇𝐶 =
40
+3
𝑄
If 𝑄 = 0, 𝐴𝑇𝐶 = ∞
If 𝑄 = ∞, 𝐴𝑇𝐶 = 3
So 𝐴𝑇𝐶 is non-linear downward
(rectangular hyperbola) has an equal
area = 40 =𝐹𝐶.
54
Chapter 5
Comparative analysis and the concept of derivative.
A comparative static analysis can be either qualitative in nature like weather
increaseor decrease in investment will increase or decrease the equilibrium
income, here weconsider the direction of the change.
The analysis will be quantitative as we consider the size of change in income
due to change in investment. (We could use the investment multiplier and
know the direction of the change from the algebraic sign.
•
Slope of a Linear Function
Rate of change and the derivative:
Suppose we have linear function 𝑦 = (𝑥)
𝑦 = 𝑎 + 𝑏𝑥
We could know the slope by locating two points (A, B) then find the
corresponding change in (y) due to change in (𝑥).
𝑦 + ∆𝑦 = 𝑎 + 𝑏(𝑥 + ∆𝑥)
∆𝑦 = 𝑎 + 𝑏𝑥 + 𝑏∆𝑥 − 𝑦
∆𝑦 = 𝑏∆𝑥
∆y
BC
= 𝑏 = 𝑠𝑙𝑜𝑝𝑒 =
∆𝑥
CA
The slope is constant and positive, the slope is thechange in (𝑦) due to change
in (𝑥).
55
•
Slope of non-linear function
The slope of the curve at a given point is the slope of tangent at this point, the
slope is not constant along the curve.
The slope at (A) =
∆y1
∆𝑥1
The slope at (B) =
∆y2
∆𝑥2
∆y1 ∆y2
≠
∆𝑥1 ∆𝑥2
This Photo by Unknown Author is licensed under CC BY-SA
•
What is derivative?
56
Derivative is a mathematical expression that measure the rate of change in (𝑦)
when (∆𝑥) is very small “close to zero”.
The derivative of a given function = (𝑥) as follows:
𝑑𝑦
∆y
≡ 𝐹(𝑥) ≡ lim
∆𝑥→0 ∆𝑥
𝑑𝑥
❖ 𝐄𝐱𝐚𝐦𝐩𝐥𝐞: 𝐼𝑓 𝑦 = 𝐹 (𝑥 ) = 3𝑥 2 − 4, 𝑓𝑖𝑛𝑑 𝑙𝑖𝑚
∆𝑦
∆𝑥→0 ∆𝑥
?
𝑦 + ∆𝑦 = 3(𝑥 + ∆𝑥)2 − 4
∆𝑦 = 3(𝑥2 + ∆𝑥2 + 2𝑥∆𝑥) − 4 − 𝑦
∆𝑦 = 3𝑥2 + 3∆𝑥2 + 6𝑥∆𝑥 − 4 − 𝑦
∆𝑦 = 3∆ (∆𝑥 + 2𝑥)
•3∆𝑥 (∆𝑥 + 2𝑥) = ∆𝑥(3∆𝑥 + 6𝑥)
∆𝑦
= 6𝑥 + 3∆𝑥
∆𝑥
∴ lim 6𝑥
∆𝑥→0
∴
𝑑𝑦
= 6𝑥
𝑑𝑥
The derivative and the slope of the curve:
Both have to do with “marginal” which used in Economics.
Example: The slope of (𝑇𝐶) curve measure (marginal) this is the change in
(𝑇𝐶), due to producing an extra unit of production the slope is geometric
explanation, thederivative of (𝑇𝐶) is the rate of change in (𝑇𝐶) due to small
increase in production.
57
∆𝑇𝐶
𝑑𝑇𝐶
= 𝑀𝐶 =
= 𝐹(𝑇𝐶)(𝑄)
∆𝑥→0 ∆𝑄
𝑑𝑄
lim
Rules of differentiation in the case of function with one independent
variable calculating derivatives by calculating:
∆𝑦
∆𝑥→0 ∆𝑥
lim
is time consuming, instead we have rules to find the derivatives of a function.
Rule :(1): The power rule
Suppose: 𝑦 = (𝑥) = 𝐾. 𝑥𝑛
𝑑𝑦
= 𝐹 (𝑥 ) = 𝑛. 𝐾. 𝑥 𝑛−1
𝑑𝑥
𝐾: 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
❖ Example:
𝑦 = −5𝑥2 ,
𝑑𝑦
𝑑𝑥
= −10𝑋. Find:
𝑑𝑦
𝑑𝑥
𝑖𝑓 𝐹(𝑥 ) = 100
(Constant function rule, 𝑛 = 0)
(𝑥) = 100𝑥0
𝑑𝑦
= 𝐹((𝑥 ) = 0(100)𝑥 −1 = 0
𝑑𝑥
𝐸𝑥: 𝐹(𝑥 ) = 0.3 𝑥 0.5. 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
𝑑𝑦
= 0.5 . 0.3 𝑥 0.5−1 = 0.15𝑥 −0.5
𝑑𝑥
Rule: (2): The generalized power rule
If (𝑥) = [(𝑥)], Then
𝑑𝑦
𝑑𝑥
= 𝐹 (𝑥 ) = 𝑛 [G(x)]𝑛−1. 𝐺(𝑥)
58
❖ Example: (𝑥) = [𝑥3 + 2]4
𝐹́ (𝑥) = 4[𝑥3 + 2]3. 3𝑥2
Rule: (3): Derivative of sums, differences of power functions
𝑑 [F(x) ± G(x)]
= 𝐹 (𝑥 ) + 𝐶(𝑥)
𝑑𝑥
❖ Example: (𝑥) = −1125 + 250𝑥 − 5𝑥2
𝐹́ (𝑥) = 250 − (2)(5) 𝑥 = 250 − 10𝑥
Rule: (4): Derivative of a product of functions
𝑦 = 𝐹(𝑥) = 𝐺(𝑥). 𝐻(𝑥)
𝐹́ (𝑥) = 𝐺(𝑥). 𝐻́ (𝑥) + 𝐻(𝑥). 𝐺́ (𝑥)
❖ Example: 𝑦 = (𝑥) = (6𝑥2 + 5𝑥). (2𝑥4 − 10)
𝐹́ (𝑥) = (6𝑥2 + 5𝑥). 8𝑥3 + (2𝑥4 − 10). (12𝑥 + 5)
𝐹́ (𝑥) = 48𝑥5 + 40𝑥4 + 24𝑥5 − 50 − 120𝑥 + 10𝑥4
𝐹́ (𝑥) = 72𝑥5 + 50𝑥4 − 120𝑥 – 50
Rule: (5): Derivative of the quotient of function
If: 𝑦 = 𝐹 (𝑥 ) =
𝐺(𝑥)
𝐻(𝑥)
𝐹 (𝑥 ) =
❖ Example: 𝑦 =
𝐹 (𝑥 ) =
𝐻 (𝑥 ). 𝐺 (𝑥) − 𝐺 (𝑥 ). 𝐻 (𝑥)
[𝐻 (𝑥)2]
6𝑥 2 +5𝑥
2𝑥 4 −10
(2𝑥 4 − 10). (12𝑥 + 5) − (6𝑥 2 + 5𝑥). (8𝑥 3 )
[2x4 − 10]2
59
−24𝑥 5 − 30𝑥 − 120𝑥 − 50
𝐹(𝑥 ) =
[2𝑥4 − 10]2
Rule: (6): Derivative of a function of a function (the chain rule)
If 𝑦 = (𝑧), 𝑧 = (𝑥)
𝑑𝑦 𝑑𝑦 𝑑𝑧
=
=.
= 𝐹 (𝑧). 𝐺(𝑥)
𝑑𝑥 𝑑𝑧 𝑑𝑥
❖ Example:
𝑦 = 2𝑧2 , 𝑧 = 3𝑥 + 2, 𝐹𝑖𝑛𝑑:
𝑑𝑦
𝑑𝑥
By substitution:
𝑦 = 2(3𝑥 + 2)2 = 2(9𝑥2 + 4 + 12𝑥)
𝑦 = 18𝑥2 + 8 + 24𝑥
𝑑𝑦
= (18)(2)x + 24 = 36x + 24
𝑑𝑥
Chain rule
𝑑𝑦
= 4𝑧,
𝑑𝑧
𝑑𝑧
== 3
𝑑𝑥
𝑑𝑦 𝑑𝑦 𝑑𝑧
=
=.
= 4(3𝑥 + 2). 3 = 36𝑥 + 24
𝑑𝑥 𝑑𝑧 𝑑𝑥
•
Non-differentiable functions:
If the function is not continuous, where the function takes a jump at (𝑥0) Or
continuous but there is a definite (kink) at (𝑥0), in the two cases we cannot
draw a unique tangent line.
60
Using derivatives in economic analysis:
In this section we show how derivatives can be used in economic modeling
andeconomic analysis.
Economic analysis is concerned with relationships between variables and in
particular the changes that occur in one variable when another variable
changes, thederivative is concerned with the same, so we can use derivatives
to explain such economic relationships.
61
A)Measuring the point elasticity:
For a demand function (𝑄 = (𝑃)), the price elasticity of demand is defined as
∆𝑄 ∆𝑃 ∆𝑄 𝑃
∆𝑄 𝑃
/
=
.
=
.
𝑄 𝑃
𝑄 ∆𝑃 ∆𝑃 𝑄
if the change in price (𝑃) is very small.
∆𝑄 𝑑𝑄
=
∆𝑃→0 ∆𝑃
𝑑𝑃
lim
The point price elasticity of demand
𝐸𝑑 =
𝑑𝑄 𝑃
.
𝑑𝑃 𝑄
the absolute value of elasticity at a particular point.
•The demand is elastic, unit elastic and inelastic at a point where |𝑬𝒅| ⋚ 𝟏.
A) Linear demand function:
𝑄 = 𝑎 − 𝑏𝑃
𝐸𝑑 =
𝑑𝑄 𝑃
.
𝑑𝑃 𝑄
•𝑎, 𝑏 > 0
𝑑𝑄
= −𝑏
𝑑𝑃
𝐸𝑑 =
−𝑏. 𝑃
𝑃
= (−𝑏).
(𝑎 − 𝑏𝑃)
𝑄
❖ Example: Suppose, 𝑄 = 100 − 2𝑃. Find 𝐸𝑑 at any point?
𝐸𝑑 =
(−2). 𝑃
−𝑃
=
100 − 2𝑃 50 − 𝑃
62
•𝑃 ≤ 50
When P = 25, 𝐸𝑑 =
−25
= −1, |𝐸𝑑 | = 1
50 − 25
When P = 30, 𝐸𝑑 =
−30
= −1.5, |𝐸𝑑 | = 1.5
50 − 30
Demand is elastic |𝐸𝑑| > 1, for (25 < 𝑃 < 50)Demand is inelastic|𝐸𝑑| < 1, for
(0 < 𝑃 < 25)
|𝐸𝑑| = 0, when 𝑃 = 0
|𝐸𝑑| = ∞, when 𝑃 = 50, 𝑄 = 0
Price elasticity of demand is not
constant along the linear demand
curve, the higher the price the higher
the |𝐸𝑑| and vice versa.
What is |𝑬𝒅|, 𝒘𝒉𝒆𝒏 𝒅𝒆𝒎𝒂𝒏𝒅 𝒄𝒖𝒓𝒗𝒆 𝒊𝒔 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒐𝒓 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍???
B) Non-linear demand function: demand curves with constant price
elasticity
Suppose:
𝑄 = 𝑎. 𝑃 −𝑏 =
𝑎
𝑃𝑏
𝑎 is constant, 𝑏 is constant > 0.
|𝐸𝑑 | =
𝑑𝑄 𝑃
.
𝑑𝑃 𝑄
63
𝑑𝑄
= (𝑎)(−𝑏)𝑃 −𝑏−1
𝑑𝑃
(𝑎)(−𝑏)𝑃 −𝑏−1 . 𝑃 −𝑎𝑏𝑃 −𝑏−1−1−𝑏
|𝐸𝑑 | =
=
= −𝑏
𝑎. 𝑃 −𝑏
𝑎
Rectangular hyperbola demand curve unit elastic demand:
•𝑏 = −1
𝑄 = 𝑎. 𝑃 −𝑏 =
|𝐸𝑑 | =
𝑑𝑄 𝑃
−𝑎
𝑑𝑃 𝑄
𝑃2
. =
.
𝑃
𝑎𝑃−1
=
−𝑎𝑃
𝑎𝑃
𝑎
𝑃
= −1
Notice:
That 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑝𝑒𝑛𝑑𝑖𝑡𝑢𝑟𝑒 = 𝑃. 𝑄 = 𝑎 is constant at any point or price
along theunit elastic demand curve.
C) Other types of demand elasticity:
•Cross-Elasticity of demand:
Suppose 𝑄𝑑𝑥 = (𝑃𝑦)
64
|𝐸𝑐 | =
𝑑𝑄𝑑𝑥 𝑃𝑣
.
𝑑𝑃𝑣 𝑄𝑑𝑥
• (𝑥, 𝑦) 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒𝑠 |𝐸𝑐| > 0
• Income elasticity of demand:
𝑄𝑑𝑥 = 𝐹(𝐼)
|𝐸𝐼 | =
𝑎𝑄𝑑𝑥 𝐼
.
𝐷𝐼 𝑄𝑑𝑥
•𝑥 is normal good |𝐸𝐼| > 0
•𝑥 is inferior good |𝐸𝐼| < 0
• Elasticity of Supply: price elasticity: Suppose: 𝑄𝑠𝑥 = (𝑃𝑥)
|𝐸𝑠 | =
𝑑𝑄𝑠𝑥 𝑃𝑥
𝑑𝑥 𝑄𝑠𝑥
> 0, 𝑠𝑢𝑝𝑝𝑙𝑦 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑢𝑝𝑤𝑎𝑟𝑑 𝑠𝑙𝑜𝑝𝑖ng.
65
Chapter 6
Using derivatives in economic analysis
1) Analyzing revenue:
Suppose The demand function is Q = F(P) Its inverse is P = G(Q).TR=P. Q.
TR= G(Q). F(P).
Marginal revenue MR: Is the first derivative of TR with respect to Q.
𝑀𝑅 =
𝑑𝑇𝑅
𝑑𝑞
𝑀𝑅 = 𝑃 + 𝑄.
𝑑𝑝
𝑑𝑞
this equation gives the general relationship between MR, P and the slope
of theinverse of demand function.
𝑑𝑝
•In the case of Monopoly demand is downward slopping ( ) < and MR<P,
𝑑𝑞
the MR curve is below the demand curve (AR).
•In the case of perfect competition, the firm is a price taker, it can sell all
what isproduced at the same market price. the demand curve is horizontal and
P = MR.
The two curves are identical.
Price Elasticity and revenue:
𝑀𝑅 = 𝑃 + 𝑄.
𝑀𝑅 = 𝑃[1 +
1
𝐸𝑑
𝑑𝑝
𝑑𝑞
, 𝑀𝑅 = 𝑃 [1 +
𝑞. 𝑑𝑝
]
𝑝𝑑𝑞
] is negative. We have the following relationship between
MR and 𝐸𝑑 as follow:
66
MR is positive when demand is elastic, and zero when demand is unit elastic
and negative when demand is in-elastic.
2)Relationship between MC and ATC:
Suppose the cost function TC = c (Q)
𝐴𝑇𝐶 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
The
𝑄=
rate
𝑑𝐴𝑇𝐶
𝑑𝑄
of
𝑑𝑐(𝑄)
=
𝑄
change
=
𝑇𝐶 𝑐(𝑄)
=
,𝑄 > 0
𝑄
𝑄
in
[ 𝑄.𝑐`(𝑄)−𝑐 (𝑄).(1]
𝑄2
ATC respect
=
1
[𝑄]
[𝑐`(𝑄 ) −
to
𝑐(𝑄)
1
𝑄
𝑄
] = [𝑀𝐶 − 𝐴𝑇𝐶]
For (Q) > 0: when ATC is > MC, the ATC curve will be downward slopping.
When ATC is < MC, the ATC curve will be upward slopping, when ATC=
MC theATC curve reaches minimum.
Ex: suppose TC = 60Q - 12 𝑄2+ 𝑄3. find MC and ATC functions? calculate
bothMC and ATC when ATC reach minimum? Find minimum MC?
𝑀𝐶 =
𝑑𝑇𝐶
= 60 − 2(12)Q + 3 Q2
𝑑𝑄
𝐴𝑇𝐶 =
Minimum ATC:
(Quadratic function ).
𝑇𝐶
= 60 − 12 Q + Q2
𝑄
𝑑𝐴𝑇𝐶
= −12 + 2 Q = 0, Q = 6
𝑑𝑄
ATC = 60 -12 (6) + (6 X 6) =24, MC = 60 – 24 (6) +3 (6X6) = 24
Minimum MC:
𝑑𝑀𝐶
= −24 + 6Q = 0,6Q = 24, Q = 4
𝑑𝑄
MC = 60 -24 (4) + 3 (4 X4) = 12
67
This solution can be graphed as follow:
Assignment:
1- Suppose TC = 100 + 2Q -2 𝑄2+𝑄3. Calculate MC function and say under
what conditions the MC function is economically meaningful?
2-Suppose TC -10 = 2 Q + 0.1 𝑄2
A- Find the expression of AC function?
B- For what values of Q will AC be decreasing? increasing? minimum?
3)Analyzing Production:
Relation between APL and MPL In the short run when capital is fixed.
̿). marginal product of labor
Assuming the production function: Q= f(L), (K=𝐾
(MPL) IS the change in total product (Q) due to very small change in labor
input assuming fixed capital.
𝑀𝑃𝐿 =
𝑑𝑄
. this is the first derivative of (Q) respect to(L).
𝑑𝐿
According to the law of diminishing returns (MPL) Is increasing at first
then decreasing eventually.
68
The slope of (MPL) function will be positive at first then become negative.
•APL: is the average product of labor. APL=
𝑂
𝐿
. it measures the total product
per unit of labor.
The relation between MPL and APL:
𝐴𝑃𝐿 =
𝑂
𝐿
, 𝑄 = 𝑓(𝐿)
𝑑𝐴𝑃𝐿
1
0
= [(𝐿). 𝑓`(𝐿) − 𝑄 (1)]/𝐿2 = 0 = (𝑓`(𝐿) − ) = 0 (for L > 0 )
𝑑𝐿
𝐿
𝐿
𝑑𝐴𝑃𝐿
≷0
𝑑𝐿
•IF MPL > APL When APL is increasing then MPL is above APL.
•IF MPL = APL When APL is maximum then MPL =APL.
•IF MPL < APL When APL is decreasing MPL is below APL.
A) Relationship between MRPL and VMPL:
Suppose Q = f (L),
TR =g(Q)
MRPL: is the rate of change in (TR) due to very small change in(L).
𝑀𝑅𝑃𝐿 =
𝑑𝑇𝑅 𝑑𝑇𝑅 𝑑𝑄
=
.
= 𝑀𝑅. 𝑀𝑃𝐿
𝑑𝐿
𝑑𝑄 𝑑𝐿
Notice in perfect competition market MR = PVMPL =P. MPL= MRPL
In monopoly market MR < P , so MRPL<VMPL.
4)Relationship between marginal product (MPL)and marginal cost
(MC):
Suppose Q =f(L), L is the only variable input assuming K is fixed. Suppose
wage rate (W), TVC =g (L) if W is given TVC = W.L
MCL is the change in TVC due to employment of an extra unit of (L) which is
69
given by: 𝑀𝐶𝐿 =
𝑑𝑇𝑉𝐶
𝑑𝐿
= 𝑊.
MC is the change in TC or TVC due to increasing output by extra unit.
𝑀𝐶𝐿 =
𝑀𝐶 =
𝑀𝐶𝐿
𝑀𝑃𝐿
=
𝑊
𝑀𝑃𝐿
𝑑𝑇𝑉𝐶 𝑑𝑇𝑉𝐶 𝑑𝑄
=
.
𝑀𝐶. 𝑀𝑃𝐿.
𝑑𝐿
𝑑𝑄 𝑑𝐿
since W is constant, as MPL increases MC decrease and
vice-versa. When MPL IS maximum MC is minimum.
5) Income tax rate and Income Multiplier:
Suppose: C = a + bYd , Yd = Y- T, T = t.Y
Equilibrium Income: Y = C + I +G
Y= a+ bYd + I+G
=a +b (Y- ty) + I + G = a + by (1-t) + I + G
Y- b.Y (1-t) =a + I + G then, Y =
𝑎+𝐼+𝐺
1−𝑏(1−𝑡)
𝑑𝑌
𝑑𝑌
1
𝑑𝐼
𝑑𝐺
1−𝑏(1−𝑡)
(𝐼 )𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑟 ( ) 𝑎𝑛𝑑 𝐺 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑟 ( ) =
The effect of change of income tax rate on K:
−𝑏
(1−𝑏+𝑏𝑡)2
𝑑𝐾
𝑑𝑡
=
=𝐾
(1−𝑏+𝑏𝑡)(0)−1(𝑏)
(1−𝑏+𝑏𝑡)2
=
<0
As (t) increases K decreases and vice versa.
6)Applying the concept of derivative on consumptionand saving
functions:
Consider the following consumption: C= f (Y) where C is the level of
consumption. Y is the level of income in the economy.
𝑀𝑃𝐶 =
∆𝑐
∆𝑦
=
𝑑𝑐
𝑑𝑦
=
𝑓`(𝑦)it is defined as the change in consumption per unit change in
income.
70
𝑀𝑃𝑆 =
∆𝑆
∆𝑌
𝑂𝑟
𝑑𝑠
𝑑𝑦
= 𝑔`(𝑦)it is defined as the change in saving per unit
change in income.
𝑀𝑃𝐶 + 𝑀𝑃𝑆 = 1
Y=C+S, c/Y+S/Y=1
Assignment:
1-Suppose the production function is given by: Q = -0.05 𝐿3+ 3𝐿2 with
0 ≤L≤ 50.
• Derive MPL and APL functions?
•Determine at what level of labor MPL = APL?
•MPL is maximum?
•TP is maximum?
•MPL equal zero?
2- Consider the following consumption function : C = f(Y) = 100 + 2𝑌0.5
•Draw the consumption function?
•Derive the expression for MPC? Comment on the behavior of MPC as
(Y) increases?
•Derive the expression for MPS? Comment on the behavior of MPS as (Y)
increases?
•Show that MPC +MPS =1?
71
Chapter 7
Partial derivatives and application to comparative static analysis
1)The concept of Partial derivatives
Economic analysis is frequently concerned with functions involving more
than two variables.
We have considered before the derivatives of function of a single independent
variable in this section. We will learn how to find the derivatives of function of
morethan one variable. (Concept of partial derivatives)
Suppose : Y =f ( , 𝑿𝟐 , ………, 𝑿𝒏 )
variables 𝑿𝒊 (I = 1, 2 ....................... n) are all independent variables.
If (𝑿𝟏) changes by 𝑿𝟏 while all other variables remain fixed, there will be
corresponding change in Y by Y + ∆ Y.
∆Y +Y =f (X𝟏 + ∆X𝟏, X𝟐, …… X𝒏)
∆Y = f (X𝟏 +∆X𝟏, X𝟐 , …………𝑿𝒏) -- f(X𝟏, X𝟐, ……. X𝒏)
dividing both sides by∆ 𝑿
∆𝑦
∆𝑥1
= 𝑓(𝑥1 + ∆𝑥1, 𝑥2 , … … . . 𝑥𝑛 ) − 𝑓(𝑥1 , 𝑥2 … … . )/∆𝑥1
∆ If we take the limit of
∆𝑦
∆𝑥1
as ∆ x1→0 , it is called the partial derivative of
(Y) with respect to (x1) holding all other variables are held constant and it is
written as
∂𝑦
∂𝑥1
or f1. the process of taking partial derivatives is called partial
differentiation.
72
2)Techniques of partial differentiation
When the function has (n) independent variables we have (n) partial
derivatives, wemust hold (n-1) independent variables constant while we allow
only one variable tovary.
❖ Example(1)
Suppose: 𝑌 = 𝑓 (𝑥1 , 𝑥2 ) = 3𝑥12 + 𝑥1 𝑥2 + 4𝑥22 . 𝑓𝑖𝑛𝑑
𝜕𝑦
𝜕𝑥1
= 𝑓1?
𝜕𝑦
= 6𝑥1 + 𝑥2 . 𝑤𝑒 𝑡𝑟𝑒𝑎𝑡 𝑥2 𝑎𝑠 constant . similarly, when calculate 𝑓2
𝜕𝑥1
𝜕𝑦
=
𝑤𝑒 𝑡𝑟𝑒𝑎𝑡 𝑥1 x1as constant 𝑓2 = 𝑥1 + 6𝑥2
𝜕𝑥1
If we know the domain of the function(f) we can calculate 𝑓1 and 𝑓2. For
example, if(𝑥1 ,2) = (1 ,3)
𝑓1(1 ,3) = 6 (1) +3 =9
, 𝑓2(1 ,3) =1 +8(3) =25
❖ Example (2)
assume that: Y =f (U, V) = (U+4) (3 U + 2 V).find: 𝑓𝑢 𝑎𝑛𝑑 𝑓𝑣 𝑤ℎ𝑒𝑛 (𝑢 = 2
, 𝑣 = 1) .
𝑓=
𝜕𝑦
= (𝑢 + 4) (3) + (3𝑢 + 2𝑣)(1)
𝑢 = 𝜕𝑢
= (2 + 4) (3) + (3(2) + 2(1) = 26
𝑓=
𝜕𝑦
= (u + 4) (2) + (3u + 2v) (0) = (2 + 4)(2) + 0 = 12
𝑣 = 𝜕𝑣
•Second order partial derivatives:
It tells us how the first partial derivative is changing.consider Ex (1):
𝜕𝑓1
𝑓1 = 6𝑥1 + 𝑥2 , 𝑓11 =
6
𝜕x1
73
𝑓2 = 𝑥1 + 8𝑥2 , 𝑓22 = 8
The second order cross partial derivative gives the rate of change in the
firstderivative when the other variable.
𝒇𝟏𝟐= 1 = 𝒇𝟐𝟏 remember that always 𝒇𝟏𝟐=𝒇𝟐𝟏 for any continuous
function.
3)Economic examples on partial derivatives
Suppose the demand function is given by: Qdx =50-2Px +3Py +0.2 I
Assume I =500
, Px =2
, py =5
Qdx=50-2(2) +3(5) +0.2 (500) =161
Price elasticity:
𝐸𝑄𝑑𝑥 ,𝑃𝑥 =
𝜕𝑄𝑑𝑥 𝑃𝑥 𝜕𝑄𝑑𝑥
,
= −2 , 𝐸𝑄𝑑𝑥 = −2(2)/161 = −0.0248
𝜕𝑃𝑥 𝑄𝑑𝑥 𝜕𝑃𝑥
→ inelastic demand
Cross elasticity:
𝐸𝑄𝑑𝑥 ,𝑃𝑦 =
𝜕𝑄𝑑𝑥 𝑃𝑦
= 3(5)/161 = 0.931 → ( X and Y are substitutes)
𝜕𝑃𝑦 𝑄𝑑𝑥
Income elasticity:
𝐸𝑄𝑑𝑥,𝐼 =
𝜕𝑄𝑑𝑥 𝐼
= 0.2(500)/161 = 0.6211 → good X is normal and necessity
𝜕𝐼 𝑄𝑑𝑥
Suppose a consumer consumes X1 and X2, the utility function is given by;
𝑼 = 𝟒𝒙𝟏 𝒙𝟐 − 𝒙𝟐𝟏 − 𝟑𝒙𝟐𝟐. .Find 𝑴𝑼𝒙𝟏 and 𝑴𝑼𝒙𝟐 and show that both are
decreasing? Calculate the cross-second derivatives?
𝑴𝑼𝒙𝟏 = 𝟒𝒙𝟐 − 𝟐𝒙𝟏 , 𝑴𝑼𝒙𝟐 = 𝟒𝒙𝟏 − 𝟔𝒙𝟐
𝜕𝑀𝑈𝑥1
𝜕𝑥1
= −2 < 0,
𝜕𝑀𝑈𝑥2
𝜕𝑥2
= −𝟔 < 𝟎
(Both
marginal
utilities
are
decreasing) so thelaw of diminishing marginal utility exists.)
74
Cross second derivatives:
𝝏𝑴𝑼𝒙𝟏
𝝏𝑴𝑼𝒙𝟐
=𝟒 ,
=𝟒
𝝏𝒙𝟐
𝝏𝒙𝟏
(Both cross derivatives are equal).
4)Market Model
Consider again the linear market model.
Qd = a – bP
(a, b > 0)
, Qs= -c +dP
(c, d >0 ) The
solution using the reduced form.
𝑃=
𝑎+𝑐
,
𝑏+𝑑
𝑂=
𝑎𝑑 − 𝑏𝑐
𝑏+𝑑
To find how a change in one of the parameters will affect the value of (P)
and (Q)We need to differentiate with respect to each of the parameters.
If the sign is (+) this indicates direct relation but if the sign is (-) the
relation isopposite.
We can get the following four partial derivatives of price. We could show
theseresults graphically.
𝜕𝑃 (𝑏 + 𝑑 )(1) − (𝑎 + 𝑐)(0)
1
=
=
>0
𝜕𝑎
(𝑏 + 𝑑)2
(𝑏 + 𝑑)
𝜕𝑃 (𝑏 + 𝑑 )(0) − (𝑎 + 𝑐)(1)
(𝑎 + 𝑐)
=
=
>0
𝜕𝑏
(𝑏 + 𝑑)2
(𝑏 + 𝑑)2
𝜕𝑃
1
=
>0
𝜕𝑐 (𝑏 + 𝑑)
𝜕𝑃 (𝑏 + 𝑑 )(0) − (𝑎 + 𝑐)(!) (𝑎 + 𝑐)1
=
=
>0
𝜕𝑑
(𝑏 + 𝑑)2
(𝑏 + 𝑑)
75
Notice that:
𝜕𝑃 𝜕𝑃
=
>0
𝜕𝑎 𝜕𝑐
And
𝜕𝑃 𝜕𝑃
=
>0
𝜕𝑏 𝜕𝑑
Test yourself by finding partial derivatives of (Q) in the reduced form
with respectto ( a, b, c and d) and show your results graphically.
5)National Income Model
Consider the following simple model of closed economy with three
endogenous variables (Y, C, and T). the equilibrium condition for
national income.
Y = C +I + G, C= a +b(Y-T),
{(𝒂 > 𝟎,𝟏 >𝒃 > 𝟎}
76
̅ > 0, 1 < 𝑡> 0}
{T
T = T̅ + ty,
This model can be solved for(Y), Y = a + b (y- (T̅ + ty) +I +G
Y= a +by -bT̅ -btY + I +GY - by+ btY =a -bT̅ + I +G
̅ + 𝐈 + 𝐆, 𝐘
𝐘[1 − b + bt] = 𝐚 − 𝐛𝐓
=(
𝟏
̅̅̅ + 𝐈 + 𝐆)
) (𝐚 − 𝐛 𝐓
𝟏 − 𝒃 + 𝒃𝒕
Three important partial derivatives can be obtained from this equation and
have special policy significance.
∗
∗
𝜕𝑌
1
=
>0
𝜕𝐺 1 − 𝑏 + 𝑏𝑡
∗
𝜕𝑌
−𝑏
=
>0
𝜕𝑇 1 − 𝑏 + 𝑏𝑡
𝜕𝑦 (1 − 𝑏 + 𝑏𝑡)(0) − (𝑎 − 𝑏𝑇̅ + 𝐼 + 𝐺(𝑏)
−𝑏. 𝑌
=
=
<0
(1 − 𝑏 + 𝑏𝑡)2
𝜕𝑡
(1 − 𝑏 + 𝑏𝑡)
Test yourself. what is
𝜕𝑦
𝜕𝑏
?
Assignment:
1- If:
Y = 𝑥𝛼 + 𝑥𝑍𝛽
,
(α, β ) are constant.
•Find the first partial derivatives with respect to X and Z?
•Find the second order partial derivatives with respect to X and Z?
•Find the second order cross partial derivatives?
2-If Y =𝑋∝ +𝑍𝛽
.
(α, β are constant)
•Find the First derivatives with respect to X and Z?
77
•Find the second derivatives with respect to X and Z?
•find the second cross partial derivatives (fxz , 𝑓𝑧𝑥) ?
3-Find out whether the law of diminishing marginal utility holds, for both goods
(A and B) in the following utility functions:
•U = 85. AB – 1.6𝐴2𝐵2
• U =0.2𝐴−1. 𝐵−1+5 AB
4-Given a demand function: Q =A 𝑃∝. 𝐼𝛽
(A, α, β ) are constant.
(P) refers to product price, (I) refers to consumer’s income.
•Calculate price and income elasticity of demand?
78
Chapter 8
Optimization Relative Maximum and Minimum
1) Definition of Relative maximum and minimum
The Concept of Equilibrium state in economics is defined as the optimum
position for a given economic unit (household, business firm or the entire
economy). The optimum positions can be determined using differential
calculus.
The Optimization problems could be maximizing something, such as
maximizing aconsumer’s utility, maximizing a firm’s profits. Or minimizing
something, such as minimizing the cost of producing a given output. The term
(relative maximum) and (relative minimum) are mathematical concepts.
In general, we have the objective function in which the dependent variable
representsthe object of maximization or minimization.
The optimization process is simply to
find the set of values of choice
variable that will give the desired
maximum or minimum.
For example, the business firm is
seeking to maximize profits. That is
maximizing the difference between
TR and TC, both are a function of (Q)
, The optimization problem is to
choose the level of (Q) that make
profits maximum, (Q) is the choice variable.
79
In general, if the objective function is Y =f(X). We need to develop the process
for finding the level of (X) That maximize or minimize (Y) assuming the
function (f) iscontinuously differentiable. 2Condit
2)Conditions of Relative maximum and minimum
The function Y = f(X) has relative maximum at point (D) and relative
minimum at point (C). (D) is maximum relative to points around (D), and (C) is
minimum relativeto points around (C).
First order condition or (necessary) for relative maximum at (D) or relative
minimum at (c) is
𝜕𝑦
𝑑𝑥
= 0(the slope is zero) at 𝑋2 and 𝑋1 Because the tangent
line is horizontal.
The second order or sufficient condition is
𝑑2 𝑦
𝑑2 𝑥
<0 to have maximum at (D)
and >0to have minimum at (C).
Let us apply these conditions on the previous example of maximizing profits we
did before.
Necessary
condition:
𝑑2 𝜋
Sufficient condition:
𝑑𝑥 2
𝑑𝜋
𝑑𝑥
=0
=0
80
❖ Example: Find the relative maximum
or minimum of the following
function:
Y=
𝒙𝟑
- 12𝒙𝟐 + 36 x +8
First order condition: :
𝒅𝒚
𝒅𝒙
= 𝟎,
𝒅𝒚
𝒅𝒙
=
𝟑𝒙𝟐 − 𝟐𝟒𝒙 + 𝟑𝟔 = 𝟎
(3x -6) (x -6) =0.3x =6
, x =2 or x =6
𝒅𝒚
= 𝟎 𝐚𝐭 𝐱 = 𝟐𝐨𝐫 𝐱 = 𝟔
𝒅𝒙
Sufficient condition :
𝑑2 𝑦
𝑑2 𝑥
= 6𝑥 − 24
• If x = 2, the second derivative = -12 < 0 at this point the function is
maximum.
• If x = 6, the second derivative = 12 > 0. The function is minimum. Y =40
at X= 2(Maximum) and Y= 8 at X =6 (minimum).
The following graph shows these results.
3)Using Optimization in Economic analysis
Profit maximization.
Suppose TR(Q), T C (Q). the profit function π (Q)= TR (Q)
– TC(Q). to find the profit maximizing output level, we must satisfy the
firstorder and second order conditions.
First order condition:
𝑑𝜋
𝑑𝜋 𝑑𝑇𝑅 𝑑𝑇𝐶
= 0,
=
=
= 0 𝐢𝐟𝐌𝐑 = 𝐌𝐂.
𝑑𝑄
𝑑𝑄
𝑑𝑄
𝑑𝑄
81
Second order condition:
𝑑 2𝜋
𝑑𝑀𝑅 𝑑𝑀𝐶
𝑑𝑀𝑅
𝑑𝑀𝐶
=
=
<
0
𝐢𝐟
<
𝑑𝑄 2
𝑑𝑄
𝑑𝑄
𝑑𝑄
𝑑𝑄
This means the rate of change in MR (the slope) is less than the rate of change
in MC (the slope). this will happen when MC is increasing (TC must
increasing at increasing rate) and MR is decreasing .these conditions can be
shown graphically as follow. the level ofproduction that maximize profit is 𝑄3.
All other levels of production are not maximizing profit.
82
(𝑸𝟐) 𝒂𝒏𝒅 (𝑸𝟒) make zero profit.
(𝑸𝟏) make loses even MR=MC, at this level the necessary condition only
is satisfied.
(𝑸𝟓) TC >TR we have losses.
❖ Example
Let us return to the partial market model assuming an excise tax as (t) per unit
ofsales.
Consider linear demand and supply functions:
Qd =a-bp, Qs= -C +d(P-t)(a , b ,c ,d>0) The equilibrium condition: Qd = Qs
after imposing excise tax (t).
83
The reduced form for equilibrium quantity:
𝑄=(
𝑎𝑑−𝑏𝑐
𝑏+𝑑
)−(
𝑏𝑑
𝑏+𝑑)𝑡
) 𝑡, for we introduce a new function of tax revenue
government. T = t.𝑄 , (T) refer to tax revenue.
(t) refer to a tax per unit ofsales.
𝑇=𝑡 (
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
)−(
) 𝑡2
𝑏+𝑑
𝑏 + 𝑑)𝑡
We require the value of (t) that will generate the maximum value of (T). , the
first order condition :
𝑑𝑇
𝑑𝑡
=0
The second order condition:
𝑑2 𝑇
𝑑𝑡 2
<0
𝑑𝑇
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
=(
) − 2𝑡 (
) = 0 → necessary condition
𝑑𝑡
𝑏+𝑑
𝑏+𝑑
(
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
𝑎𝑑 − 𝑏𝑐
) = 2𝑡 (
) ∴ 𝑎𝑑 − 𝑏𝑐 = 2𝑡 (𝑏𝑑 ) → 𝑡 =
𝑏+𝑑
𝑏+𝑑
2𝑏𝑑
𝑑 2𝑇
𝑏𝑑
=
−2
(
) < 0 → sufficient condition
𝑑𝑡 2
𝑏+𝑑
❖ Example:consider the following demand and supply functions:
𝑄𝑑 = 40 − 2𝑃
, 2P - Qs = 20
𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑡ℎ𝑒 𝑒𝑥𝑖𝑠𝑒 𝑡𝑎𝑥 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑠𝑎𝑙𝑒𝑠 𝑖𝑠 (𝑡), 𝑓𝑖𝑛𝑑:
• 𝑡ℎ𝑒 𝑡𝑎𝑥 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤ℎ𝑖𝑐ℎ 𝑤𝑖𝑙𝑙 𝑚𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑟𝑒𝑣𝑒𝑛𝑢𝑒?
•Calculate the maximum revenue that can be obtained from taxation?
•Use the derivative test to show your answer?
84
𝑄=(
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
)−(
) 𝑡, 𝑇 = 𝑡 (
)−(
)𝑡 2
𝑏+𝑑
𝑏+𝑑
𝑏+𝑑
𝑏+𝑑
𝑑𝑇
𝑎𝑑 − 𝑏𝑐
𝑏𝑑
𝑑𝑇
=(
) − 2𝑡 (
)=0 →
𝑑𝑡
𝑏+𝑑
𝑏+𝑑
𝑑𝑡
(40)(2) − (2)(20)
(2)(2)
=
2𝑡
→𝑡=5
(2 + 2)
(2 + 2)
(2)(2)
𝑑 2𝑇
𝑏𝑑
=
−2
(
)
<
0
→
−2
= −2 < 0
(2 + 2)
𝑑𝑡 2
𝑏+𝑑
the sufficient condition is satisfied for maximum.
Maximum tax revenue : T = t. Q̅
𝑄=
(40)(2) − (2)(20) (2)(2)
(5) = 5, 𝑇 = (5)(5) = 25
−
(2 + 2)
(2 + 2)
A)Maximizing Total revenue.
Suppose the demand function Q =a- b. P
(a , b >0)
𝑎
1
𝑇𝑅 = 𝑃. 𝐴, 𝑏. 𝑃 = 𝑎 − 𝑄, 𝑃 = ( ) − (𝑄 ) → 𝑡ℎ𝑒𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓
𝑏
𝑏
𝑎
1
𝑇𝑅 = ( ) 𝑄 − (𝑄 2 ), TR will be at maximum when MR = 0
𝑏
𝑏
𝑀𝑅 =
𝑑𝑇𝑅
𝑑𝑄
= 0 (The first order condition) and
𝑑2 𝑇𝑅
𝑑𝑄2
< 𝑜 (the second order
condition).
𝑀𝑅 =
𝑑 2 𝑇𝑅
𝑑𝑄2
𝑑𝑇𝑅
𝑎
2
2
𝑎
𝑎
= ( ) − ( )𝑄 = 0 → ( )𝑄 = ( ) → 𝑄 = ( )
𝑑𝑄
𝑏
𝑏
𝑏
𝑏
2
2
𝑎
𝑏
2
= − < 0 , TR is maximum if 𝑄 =
• 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒗𝒆𝒏𝒖𝒆 𝒂𝒏𝒅 𝑷𝒓𝒊𝒄𝒆 𝑬𝒍𝒂𝒔𝒕𝒊𝒄𝒊𝒕𝒚 𝒐𝒇 𝒅𝒆𝒎𝒂𝒏𝒅
85
𝐸𝐷 = −
𝑑𝑄 𝑃
𝑃
𝑎
𝑎
1 𝑎
𝑎
. = −𝑏. = −𝑏 𝑎. , 𝑄 = → 𝑃 = ( ) − ( ) ( ) 𝑃 = ( )
𝑑𝑃 𝑄
𝑄
2
𝑏
𝑏 2
2𝑏
𝑧
𝐸𝐷 = −𝑏. (
𝑎
𝑎
𝑎
2
) / = −𝑏 ( ) . ( ) = −1
2𝑏 2
2𝑏
𝑎
TR will be maximum when ED =- 1 (demand is unit elastic) and MR = 0. This
relation can be explained as before from the formula: 𝑀𝑅 = 𝑃[1 +
1
𝐸𝐷
]
B)Cost minimization.
Suppose TC = a + b.Q +c 𝑄2.
We need to find out the level of output (Q) at which the average cost (AC)
is minimum.
𝐴𝐶 =
Two
And
conditions
𝑑2𝐴𝐶
𝑑𝑄2
must
be
𝑇𝐶 𝑎
= + 𝑏 + 𝑐𝑄.
𝑄
𝑄
satisfied :
𝑑𝐴𝐶
𝑑𝑄
= 0 (necessary
condition).
> 0 (sufficient condition).
𝑑𝐴𝐶
𝑎
𝑎
= −𝑎𝑄 −2 + 𝑐 = 0 → 𝑐 = 2 → 𝑎 = 𝑐𝑄 2 → 𝑄 2 =
𝑑𝑄
𝑄
𝑐
𝑄 = ±√
𝑎
𝑐
𝑑 2𝐴𝐶
𝑎
𝑎 𝑑 2𝐴𝐶
−3
= 2𝑎𝑄 →= 3 > 0, 𝑤ℎ𝑒𝑚 𝑄 = √
=< 0
𝑑𝑄 2
𝑄
𝑐 𝑑𝑄 2
𝑀𝐶 =
𝑑𝑇𝑅
= 𝑏 + 2. 𝑐𝑄. 𝑤ℎ𝑒𝑛 𝑀𝐶 = 𝐴𝐶, 𝐴𝐶 must be minimum
𝑑𝑄
𝑏 + 2𝑐𝑄 =
𝑎
𝑎
𝑎
𝑎
+ 𝑏 + 𝑐𝑄. → 𝑐. 𝑄 = → 𝑐𝑄 2 = 𝑎 → 𝑄 2 = , 𝑄 = ±√
𝑄
𝑄
𝑐
𝑐
86
Derivation of Supply Curve.
Supply curve is defined for a price-taker firm as the upward section of MC
curveafter intersecting AVC curve at its minimum.
❖ Example:
Suppose TC function is given by:
TC= 0.1𝑸𝟑- 2𝑸𝟐 +15 Q +10. Assuming the industry consists of 100
identicalfirms.
•Derive individual and aggregate supply curve? if (P) =22.5 what is the quantity
supplied by each firm and industry?
𝑀𝐶
𝑑𝑇𝐶
𝑑𝑄
= (3)(0.1)𝑄 2 − 4𝑄 + 15, the necessary condition for maximizing
profits:
MC =MR =P
P =0.3 𝑄2-4Q+ 15.
→ 0.3 𝑄2 – 4Q +15 -P=0
(quadratic
function). Solving for Q: using the formula
−𝑏 ± √𝑏2 − 4𝑎𝑐
[a = (15 − P), b = −4 , c = 0.3]
2𝑐
(𝑄 ) = −(−4) ± √16 − 4(0.3)(15 − P) / 0.6
𝑄 = 4 + √16 − 18 + 1.2P / 0.6 = 4 + √√−2 + 1.2P/ 0.6
Finding the minimum AVC:
𝒅𝑨𝑽𝑪
= 𝟎. 𝟐𝑸 − 𝟐 = 𝟎
𝒅𝑸
𝒅𝟐 𝑨𝑽𝑪
= 𝟎. 𝟐 > 𝟎
𝒅𝑸𝟐
𝑨𝑽𝑪 = 𝟎. 𝟏𝑸𝟐 − 𝟐𝑸 + 𝟏𝟓,
, 𝑸 = 𝟏𝟎
Substitute Q =10 in AVC function. We get the minimum price to supply, below
thisprice no supply.
87
P= 0.1(100) -2(10) +15=5.
The equation of individual supply: 𝑄𝑠 = 4 + √−2 + 1.2P / 0.6
if P ≥ 5
. if P < 5 Qs = 0
The equation of aggregate supply: 𝑄𝑠 = (100)4 + √−2 + 1.2P / 0.6 IF P ≥
0 If P <5 no supply.
At P= 22.5. Qs for each firm =15 units, the industry supply =1500 units.
Assignment:
1)the market demand is given by: 4P+Q – 16 =0 and the AC function takes the
form: AC = 4 +2 -0.3 Q +0.05𝑄2
Find the (Q) which gives:(use the second derivative test in each case?
•Maximum revenue.
•Minimum marginal cost.
•Maximum profits.
2)A firm has the following total cost and demand functions:
𝑇𝐶 =
1 3
𝑄 − 7 𝑄 2 + 111 𝑄 + 50, 𝑄 = 100 − 𝑃
3
Find the profit maximizing level of output? Calculate the maximum profits?
3)Let the TR(Q) =1200 Q-2𝑄2, TC =𝑄3- 61.25𝑄2 +1528.5 Q+2000
Find the output that maximize the profits and calculate these profits?
88
Chapter 9
Total differentials and total derivatives
1)Total differential
In the case of function of one variable 𝒀 = 𝒇(𝒙),
𝒅𝒚
𝒅𝒙
𝑳𝒊𝒎
∆𝒚
∆𝒙
𝒘𝒉𝒆𝒏 ∆𝒙 → 𝟎.
𝒅𝒚
= 𝒇ˊ(𝒙) → 𝒅𝒚 = 𝒇ˊ(𝒙). 𝒅𝒙 is called differential. It gives the change in Y
when the change in X is very small approaches to zero.
𝒅𝒙
𝑻𝒉𝒆 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 (𝒚) = 𝒕𝒉𝒆 𝒇𝒊𝒓𝒔𝒕 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒐𝒇 (𝒚)𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐(𝑿). times
the change in (X). we could apply the differential concept to functions of more
thanone independent variable, this is called total differential.
, ………….𝑿𝒏 )
If Y = F(X1, X2 ,
𝑑𝑌 =
𝜕𝑓
𝜕𝑥1
𝑑𝑥1 +
𝜕𝑓
𝜕𝑥 2
𝑑𝑥2 … … … . .
𝜕𝑓
𝜕𝑥𝑛
𝑑𝑥𝑛 , (𝑑. 𝑌) give the approximate value
of the total change in (Y) resulting from very small change in each
independent variable.
❖ Example(1)
If Y = 5x12 + 3x2 . Find d Y?
𝑑𝑌 =
𝜕𝑓
𝜕𝑓
𝑑𝑥1 +
𝑑𝑥 (10𝑥1 )𝑑𝑥1 + 3𝑑𝑥2
𝜕𝑥1
𝜕𝑥2 2
❖ Example(2)
find the total differential of the function: 𝒀 = 3𝑥12 + 𝑥1 𝑥22
𝑑𝑌 =
𝜕𝑓
𝜕𝑓
2 )𝑑𝑥1 + (2𝑥1 𝑥2 )𝑑𝑥2
𝑑𝑥1 +
𝑑𝑥 (𝑥
𝜕𝑥1
𝜕𝑥2 2 1+𝑥 2
89
*Rules of differentials*
 1- dk = 0= (K is constant function)
 2- d (C𝑼𝒏 ) =nC 𝑼𝒏−𝟏.du
 3- d (U ± V) =d U±d U
𝑼
𝑽.𝒅𝑼−𝑼.𝒅 𝑽
 𝟒−𝒅𝑽 =
𝑽𝟐
 5- d (U. V) =U. d V + V.d U
2)Total derivative
We have Considered a function where (X) variables are independent of each
other.in some cases or economic models we may have a situation where Y =f
(x,w) and X =g(W). the three variables (Y, X and G) are related as shown in
the following figure. When (W) changes it has direct effect on (Y) through
function(f).
It has also indirect effect through function (g), by affecting on (X), when (X)
changes it affect (Y) through function (f).
the total effect is given by total derivatives and can be calculated using total
differential.
(𝑑𝑌)
𝜕𝑌
𝜕𝑊
𝑑𝑊 +
(
𝜕𝑌
𝜕𝑋
𝑑𝑌
)
𝑑𝑊
total=
𝑑𝑋 by dividing both sides by d W., we get:
𝜕𝑌
+
𝜕𝑊
𝜕𝑌 𝑑𝑋
𝜕𝑋 𝑑𝑊
↓
↓
direct effect
indirect effect
90
❖ Example
𝐼𝑓 𝑌 = 𝑓 (𝑌) = 6𝑥13 + 7𝑥2 , 𝑥2 = 𝑔(𝑥1 ) = 4𝑥12 + 3𝑥1 + 8, Find
𝑑𝑦
𝑑𝑥1
(At
X1=1)
(
𝑑𝑦
𝜕𝑌
𝜕𝑌 𝑑𝑥2
)=
+
= 18𝑥12 + (7)(8𝑥1 + 3) = 95
𝑑𝑥1
𝜕𝑥1 𝜕𝑥2 𝜕𝑥1
If Y = f(Y)=6 𝑿𝟑 + 7 𝑿𝟐
,
𝑿𝟐= g(𝑿𝟏) = 4𝑿𝟐 +3𝑿𝟏+ 8
3)Some economic examples on total differentials
•Saving function:
Consider the saving function:S = (Y, I) where (S) refers to total saving,
(y) is the national income, (I) is the interest rate. assuming the function
have partial derivatives.
𝝏𝑺
( ) measure the rate of change of (S) with respect to very small change
𝝏𝒀
in(Y) assuming (I) hold constant.
𝝏𝑺
( ) = 𝑴𝑷𝑺 = 𝑺𝒀 . By the same token, the change in (S) resulting from very
𝝏𝒀
𝝏𝑺
small change in (I) holding (Y) constant is given by ( ) = 𝑺𝒊
𝝏𝒊
The total change in (S) due to change in (Y) and (I) is given by:
𝝏𝑺
𝝏𝑺
𝒅𝑺 =
. 𝒅𝒀 +
. 𝒅𝒊 = 𝑺𝒀 . 𝒅𝒀 + 𝑺𝒊 . 𝒅𝑰
𝝏𝒚
𝝏𝒊
it is possible that (Y) may change while (i) remains constant, in that case d
𝝏𝑺
𝒅𝑺
𝝏𝑺
I=0. the total differential will reduce to 𝒅𝑺 = 𝒅𝒚 𝒐𝒓
= .
𝝏𝒀
𝝏𝒀
If (Y) remain constant while (i) changes, 𝑑𝑦 = 0 & 𝑑𝑆 =
𝝏𝒀
𝜕𝑆
𝜕𝑖
. 𝑑𝑖 →
𝜕𝑆
𝜕𝑖
=
𝑑𝑆
𝑑𝑖
we consider what we call partial elasticity of s a v i n g with respect to (Y)
orrespect to (i).
𝝏𝑺 𝑺 𝝏𝑺 𝒀
÷ =
. > 𝟎 (𝐬𝐚𝐯𝐢𝐧𝐠 𝐞𝐥𝐚𝐬𝐭𝐢𝐜𝐢𝐭𝐲 𝐭𝐨 𝐢𝐧𝐜𝐨𝐦𝐞, 𝐡𝐨𝐥𝐝𝐢𝐧𝐠(𝐢)𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕)
𝝏𝒚 𝒀 𝝏𝒀 𝑺
𝝏𝑺 𝑺 𝝏𝑺 𝒊
𝑬𝒔𝒊 =
÷ =
. > 𝟎(𝐬𝐚𝐯𝐢𝐧𝐠 𝐞𝐥𝐚𝐬𝐭𝐢𝐜𝐢𝐭𝐲 𝐭𝐨 𝐢𝐧𝐭𝐞𝐫𝐞𝐬𝐭, 𝐡𝐨𝐥𝐝𝐢𝐧𝐠 (𝐘) 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭)
𝝏𝒊 𝒊
𝝏𝒊 𝑺
𝑬𝒔𝒚 =
91
•Indifference curve and calculating MRS.
Suppose total utility (U) depends on the amounts received of two goods, the
utility function is given by:U = u (x , y) .
The indifference curve is defined as the curve that show different
combinations that give the consumer the same level of utility or satisfaction
(𝑼 = 𝒖) so He is indifferent to choose.
The rate of change in(U) is zero along the indifference curve U =u (x, y) =𝑢
𝑑𝑈 =
𝜕𝑈
𝜕𝑈
d𝑋1 +
d𝑋 , since (U) is constant, set d U = 0
𝜕𝑥1
𝜕𝑥2 2
𝜕𝑈
𝜕𝑈
𝜕𝑈
𝜕𝑈
d𝑋1 +
d𝑋2 = 0 →
d𝑋1 =
. d𝑋2
𝜕𝑥1
𝜕𝑥2
𝜕𝑥1
𝜕𝑥2
𝜕𝑈
𝑑𝑋2
𝑀𝑈𝑥1
𝜕𝑋
d𝑋1 + 1 → −(slope of indifference curve) = MRS =
𝜕𝑈
𝑑𝑋1
𝑀𝑈𝑥2
𝜕𝑥2
❖ Example: Suppose Utility function is given by:
𝑼 = 𝟒𝒙𝟏 𝒙𝟐 − 𝒙𝟐𝟏 − 𝟑𝒙𝟐𝟐
•Find MUX1, MUX2? And prove they are? decreasing?
•Calculate MRS and show it is diminishing as 𝒙𝟏increases?
𝜕𝑈
= 𝑀𝑈𝑋1 = 4𝑥2 − 2𝑥1
𝜕𝑥1
.
𝜕𝑈
= 𝑀𝑈𝑋2 = 4𝑥𝟏 − 6𝑥𝟐
𝜕𝑥1
𝑀𝑈𝑋1
𝑀𝑈𝑋2
= −2 < 0 ,
= −6 < 0 (𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 ).
𝜕𝑥1
𝜕𝑥2
𝑀𝑅𝑆 =
𝑀𝑈𝑋1 4𝑥2 − 2𝑥1
=
𝑀𝑈𝑋2 4𝑥1 − 6𝑥2
=
,
𝜕𝑀𝑅𝑆
𝜕𝑥1
(4𝑥1 − 6𝑥2)(−2) − (4𝑥2 − 2𝑥1)(4)
(4𝑥1 − 6𝑥2)2
92
𝑴𝑹𝑺 =
−𝟖𝒙𝟏+𝟏𝟐𝒙𝟐−𝟏𝟔𝒙𝟐−𝟖𝒙𝟏
(𝟒𝒙𝟏−𝟔𝒙𝟐)𝟐
=
−𝟒𝒙𝟐
(𝟒𝒙𝟏−𝟔𝒙𝟐)𝟐
< 𝟎 (So MRS is diminishing) this
means that the indifference curve is convex to the origin.
•
Iso-quant and calculating MRTS:
Iso- Quant indicates all combinations of two inputs (L, K) that produce the
same level of production. the equation of Iso-Quant is given by Q=f (L, K)
𝒅𝑲
=𝒒, the slope equal ( ) which is negative because the Iso-Quant is
𝒅𝑳
downward slopping.
Since (Q) is constant along the curve →𝑑𝑄
𝜕𝑄
𝜕𝐿
. 𝑑𝐿 +
𝜕𝑄
𝜕𝐾
. 𝑑𝐾
𝜕𝑄
𝜕𝑄
𝜕𝑄
𝑑𝐾
𝑀𝑃𝐿
𝑠𝑒𝑡 𝑑𝑄 = 0 →
. 𝑑𝐿 = − +
. 𝑑𝐾 → −
= 𝜕𝐿 → 𝑀𝑅𝑇𝑆 =
>0
𝜕𝑄
𝜕𝐿
𝜕𝐾
𝑑𝐿
𝑀𝑃𝐾
𝜕𝐾
Assignment:
1-assume the following market model after imposing an excise tax.
Qd =500- 9P
, Qs =-100+ 6 (P-t).
•Find the excise tax that maximize the government tax revenue? Use the test
derivatives.
2 − 𝐼𝑓 𝑌 = 50 − 30𝑥1 + 6𝑥12 − 5𝑥2 − 10𝑥22 − 3𝑥1𝑥2
•Find d Y? when 𝑋1changes from 5 to 5.02 and 𝑋2 changes from 2 to 2.02.
3-Consider the following utility function:
𝑈 = 5𝑥1.0.5𝑥20.5 *Draw the indifference curve when U =20?
•Find each of the following and show they are decreasing:
MUX1, MUX2 and MRS. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑀𝑅𝑆 𝑤ℎ𝑒𝑛 𝑈 = 20 𝑎𝑛𝑑 𝑥1=2 𝑎𝑛𝑑 𝑥2=4
93
4-Suppose the utility function is given by: U (X, Y) = 𝑋0.25. 𝑌0.75
Assuming the consumer’s income is $100, 𝑝𝑟𝑖𝑐𝑒𝑠 𝑜𝑓 (𝑋 𝑎𝑛𝑑 𝑦 , 𝑃𝑌.)
•Derive demand function of good X and Y?
•What would happen to demand for (X) as income and , are doubled?
•Calculate 𝐸𝐷 assuming income and is constant?
•Calculate income elasticity of demand for good (X) assuming 𝑃, is constant?
•Prove that the indifference curve is convex to origin?
94
Chapter 10
Homogeneous functions and its economic applications
1) Definition and properties (general)
A function is homogenous of degree(r) if multiplication of each of its
independent variables by positive constant (𝝀) will change the value ofthe
function by (𝝀𝒓).
Y =f (𝒙𝟏, 𝒙𝟐 , ………𝒙𝒏) is homogenous of degree(r) if f(𝒙𝟏λ ,𝒙𝟐𝝀
,….𝒙𝒏λ)=𝝀𝒓(f(𝑿𝟏,𝑿𝟐 ,…..𝑿𝒏 ).
❖ Example: Given the function 𝒁 = 𝒇(𝒙, 𝒚, 𝒘) = 𝒙𝒚 + 𝟐𝒘
𝟑𝒙
𝑓 (𝑥𝜆, 𝑦𝜆, 𝑤𝜆) =
𝑥𝜆
𝑦𝜆
+
2𝑤𝜆
3𝑥𝜆
𝒙
𝟐𝒘
𝒘
𝟑𝒙
= 𝝀𝟎 ( +
) = 𝝀𝟎 (𝒁), the function is
homogenous of degree zero, this means that the change in all variables by a
given proportionhas no effect on the function.
• Properties: (general):
A) adding any constant to homogenous function make it non homogenous.
❖ Example:
Y = 𝑋1𝑋2 (this function is homogenous of degree 2)
𝑦∗=𝜆𝑥1. 𝜆𝑥2=𝜆2(𝑥1.𝑥2 )=𝜆2(Y). suppose we add constant (K) (K≠ 0). Y̅ =̅̅λ
𝑋1 𝜆𝑋2 + K =𝜆2(Y) +K≠Y= (𝑋1𝑋2).
B) the homogenous function of degree (r) has first partial derivatives homogenous
of degree (r-1).
❖ Example:
Y = 𝑋1𝑋2 (homogenous of degree 2)
95
𝜕𝑦
𝜕𝑥1
= 𝑥2 .
𝜕𝑦
𝜕𝑥2
= 𝑥1 (Both functions are homogenous of degree 1).
2) Properties of linear homogeneity
Suppose Y = F (𝑥1,x2). this function can be written as:
𝑥1
𝑥2
𝑌 = 𝑥1𝑔 ( ) 𝑜𝑟 𝑌 = 𝑥2 ℎ ( )
𝑥2
𝑥1
❖ Example:
suppose the production function Q = f (K, L) is linearly homogenous. As K
and L increases by given proportion production will change y the same
1
𝑄
𝐾
𝑘
𝐿
𝐿
𝐿
𝑙
proportion. if we multiply by ( ) we get ( ) = 𝑓 ( , 1) → 𝐴𝑃𝐿 = 𝑓 ( ),
APL does not change as (K and L) increase by a given proportion APL is
homogenous of degree of zero. (Ratio of(K) to (L) is constant).
1
𝐿
𝐾
𝐾
If we multiply by( )we get the same result 𝐴𝑃𝐾 = 𝑓 (1, ).
𝑡
𝑙
𝑄 = 𝐴𝑃𝐿. 𝐿 = 𝐿. 𝑓 ( ) . 𝑜𝑟 𝑄 = 𝐴𝑃𝐾. 𝐾 = 𝐾. 𝑓 ( )
𝑙
𝑘
3) Homogeneity of Cobb-Douglas function
𝛽
The general form of Cobb-Douglass: 𝑌 = 𝐴𝑋1𝛼 . 𝑋2 , ( ∝ 𝑎𝑛𝑑 𝛽 are positive
constant). The function is homogenous of the degree (∝ +𝛽). the function has
manyapplications in microeconomics.it could be used as utility or production
function.
𝛽
𝛽
𝛽
𝑌 = 𝐴(𝜆𝑋1 )𝑎 . (𝜆𝑋2 ) = 𝜆 (𝐴. 𝑋1𝛼 𝜆𝑞 . 𝑋2 𝜆𝛽 ) = 𝜆𝛼+𝛽 (𝐴𝑋1𝛼 𝑋2 )
𝑌 = 𝜆𝛼+𝛽 (𝑌)
96
•If (∝ +) > 1, we have increasing returns to scale. Doubling inputs for
example cause more than double in output.
•If (∝ 𝛽) =1, we have constant returns to scale. Doubling inputs for example
causes doubling output.
•If ∝ +) < 1, we have decreasing returns to scale. Doubling inputs causes
less than double in output.
❖ Example:
suppose the production function is Q = A.𝐿∝. 𝐾𝛽 (A>0 and >1 ∝,>0).
4) Relation between average and marginal product of inputs in CobbDouglass production function
𝑸 𝑨. 𝑳∝ . 𝑲𝜷
𝑨𝑷𝑳 = =
= 𝑨𝑳∝−𝟏 . 𝑲𝜷
𝑳
𝑳
𝝏𝑸
𝑴𝑷𝑳 =
=∝. 𝑨𝑳∝−𝟏 . 𝑲𝜷 , = ∝ (𝑨𝑷𝑳) → 𝑴𝑷𝑳 < 𝑨𝑷𝑳
𝝏𝑳
*Relation between APK and MPK*
𝑄 𝑨. 𝑳∝ . 𝑲𝜷
𝐴𝑃𝐾 = =
= 𝑨𝑳∝ . 𝑲𝜷−1
𝐾
𝑳
𝑀𝑃𝐾 =
𝜕𝑄
= 𝐴. 𝛽𝐿∝ 𝐾𝛽−1 = 𝛽(𝐴𝑃𝐾) → 𝑀𝑃𝐾 < 𝐴𝑃𝐾
𝜕𝐾
*How MPL changes as L changes, and how MPK changes as K changes? Test
yourself and show they are decreasing? Think about
𝜕𝑀𝑃𝐿 𝜕𝑀𝑃𝐾
𝜕𝑄
,
𝜕𝑄
97
5) Euler’s theorem and distribution of output
The theory states that If (Y =f (𝑋1, 𝑋2 ) function is linearly homogenous it
can be expressed as the sum of terms each of which is the product of one
independent variable and the first partial derivative with respect to that
variable.
𝑌 = 𝑋1. 𝜕𝑌 + 𝑋2. 𝜕𝑌 The proof can be shown by this example.
𝜕𝑋1
𝜕𝑋2
❖ Example:
suppose :Q = A𝐿∝. 𝐾1−∝ ,
(>1 ∝ >0).
𝜕𝑄
=∝ 𝐴𝐿∝−1 ,
𝜕𝐿
𝐿.
𝐾.
𝜕𝑄
= (1−∝)𝐿∝ . 𝐾1−∝−1
𝜕𝐾
𝜕𝑄
=∝ 𝐴𝐿∝−1. 𝐾1−∝ . 𝐿 =∝ 𝐴 𝐿∝ . 𝐾1−∝
𝜕𝐿
𝜕𝑄
= (1−∝)𝐿∝ . 𝐾1−∝−1. 𝐾 = (1−∝)𝐴𝐿∝ . 𝐾1−∝
𝜕𝐿
𝐿.
𝜕𝑄
𝜑
+ 𝐾.
=∝ 𝐴𝐿∝ . 𝐾1−∝ + (1−∝)𝐿∝ . 𝐾1−∝
𝜕𝐿
𝜕𝐾
= 𝐴 𝐿∝ . 𝐾1−∝ (∝ +1-∝ ) =Q (1)
6) Production partial elasticities and relative shareof labor and capital
Consider the production function Q = f (L, K) , Q=A𝐿∝𝑲𝜷 , the elasticity of
production with respect to (L) for example is given by:
𝐸𝐿 =
𝜕𝑄 𝐿
𝐿
𝐿
. =∝ 𝐴𝐿∝−1. 𝐾𝛽. =∝ 𝐴𝐿∝ 𝐾𝛽. =∝.
𝜕𝐿 𝑄
𝑄
𝑄
•The elasticity of production with respect to (K) =𝐸𝐾 −
𝜕𝑄 𝐾
.
𝜕𝐾 𝑄
98
= 𝛽𝐴𝐿∝ . 𝐾
𝛽−1.
𝐾
𝑄
= 𝛽𝐴𝐿∝ . 𝐾
•Relative share of labor to total output =
𝐿𝑀𝑃𝐿
•Relative share of Capital to total output =
•Relative share of labor to capital =
∝
𝛽
𝐿
=
𝑄
1
𝑄
=𝛽
∝ 𝐴𝐿∝−1. 𝐾𝛽 =∝
𝑄
𝐾.𝑀𝑃𝐿
𝑄
𝛽.
=
𝐾𝛽𝐴𝐿∝ .𝐾𝛽−1
𝑄
=𝛽
if ∝ +𝛽 =1→ 𝛽=1-∝
•The relative share does not change as the amount of labor or capital it will
equal =
∝
1−∝
7) Expansion path for linear homogenous Cobb-Douglas.
Consider the production function Q = A𝑳∝𝑲𝟏−
∝
,
(1 >∝ > 0)
The Expansion path is the locus of tangency points between the Iso-cost line
and the Iso-Quant. the necessary condition is:
The slope of Iso-cost line=the slope of the Isoquant
−
𝑤 𝑑𝑘 𝑤
𝑑𝑘 𝑤
=
→ =−
→ = 𝑀𝑅𝑇𝑆
𝑟
𝑑𝑙
𝑟
𝑑𝑙
𝑟
(w and r are prices of labor and Capital, we assume they are given), MRTS
𝑀𝑃𝐿
is the marginal rate of technical substitution) which is equal to
𝑀𝑃𝐾
∝−1
∝−1
𝑀𝑃𝐿 =∝ 𝐴𝐿
=
.𝐾
1−∝
∝
, 𝑀𝑃𝐾 = (1−∝)𝐴𝐿 . 𝐾
−∝
∝ 𝐴𝐿 . 𝐾1−∝
→=
(1−∝)𝐴𝐿∝ . 𝐾 −∝
∝ 𝐾
( )
1−∝ 𝐿
𝑤
∝ 𝐾
(1−∝)𝑤𝑙 (1−∝) 𝑤
=
( ) →∝ 𝑘𝑟 = 𝑤𝑙(1−∝)𝑘 =
=
.𝐿
𝑟 1−∝ 𝐿
∝𝑟
∝ 𝑟
This the equation of the Expansion path which is straight line with positive
constant slope
(1−∝) 𝑤
∝
𝑟
> 0, the following graph show the Expansion path.
99
Assignment:
1-assume a production function of the firm is:
Q =10 𝐿0.75𝐾0.25
•Find MPL and MPK functions and prove they are positive and decreasing?
•Find the total differentials?
•Find MRTS function and show it is positive and decreasing?
•Calculate the Elasticity of production with respect to (L and K).
•What case of returns to scale in this firm?
2-Use Euler, s Theorem to show in the case of constant returns to scale production
function.
a-when KPK=0, APL is equal to MPL.
b-when MPL =0, APK is equal to MPK.
100
3- Given
the
production
function
Q
=A𝐿∝𝐾𝛽
.show
that:
a-∝ +𝛽=1→implies C.R.S.
b-∝ +𝛽 > 1 → implies I.R.S.
c-∝ +𝛽 < 1 → implies D.R.S.
d- ∝ 𝑎𝑛𝑑 𝛽are partial Elasticity of production with respect to (L and K).
𝑘
e- MPL is a function of ( )alone when (∝ +𝛽=1).
𝑙
𝑘
f- APK is a function of ( ) alone when (∝ +𝛽=1).
𝑙
101
Chapter 11
Constrained optimization and lagrange multiplier model
1) free and constrained optimization
We discussed before how to find the relative maximum or minimum of a
function without constraints (free optimization).
In this section we explain to find the optimization of the function when it
has aconstraint (constraint optimization).
Suppose a firm has a constraint or restriction in the form (𝑄1+𝑄2) =1000. The
profit maximization level of output is constrained, (t h e higher 𝑄1 the
lower, 𝑄2) .
We have many examples of constrained optimization; the consumer is
seeking to maximize utility subject to his budget constraint (his income and
good’s prices).
The firm is seeking to maximize the production subject to iso-cost or
minimizing cost subject to production constraint.
2) Finding the Optimum values assuming equality constraint
❖ Example
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑎 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑟 𝑤𝑖𝑡ℎ 𝑠𝑖𝑚𝑝𝑙𝑒 𝑢𝑡𝑖𝑙𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒
𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑢𝑡𝑖𝑙𝑖𝑡𝑦 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑤𝑜
𝑈 = 𝑋1𝑋2 + 2𝑋1
(1) 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Suppose the budget constraint is expressed by the linear equation.
60 = 4𝑥1+ 2𝑥2
(2) → income=60, 𝑝𝑥1 =4 ,2 =2
The problem now is to maximize (1) subject to constraint (2).
102
What amounts of 𝑿𝟏and 𝑿𝟐 should be purchased?
You could solve this problem using three ways.
a-substituting the constraint into the objective function: 2𝑥2 = 60- 4𝑥1 →
𝑥2= 30-2𝑥1
Combine the constraint with objective function, you get function of one
variable.
= 32 𝑋1 -2𝑋12
U =𝑋1(30-2𝑥1)+ 2𝑋1
To maximize utility, the first order condition is
𝑋1∗ = 8
,
𝜕𝑈
𝜕𝑈1
− 32 − 4𝑥1 = 0
2
𝜕𝑈
2 <
𝜕𝑋1
𝑡ℎ𝑒 𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:
2
0, −𝜕𝑈 = − 4 < 0
2
𝜕𝑋1
(sufficient condition issatisfied for maximum)
𝑋2∗ =
60 − 4(8)
= 14, 𝑈 ∗ = (8)(14) + 2(8) = 128
2
b-graphical solution:
At the tangency point:
𝑀𝑅𝑆 =
𝑀𝑢𝑥1 =
𝑝𝑥 𝑀𝑢𝑥1
=
𝑝𝑦 𝑀𝑢𝑥2
𝜕𝑈
𝜕𝑈
= 𝑥2 + 2, 𝑀𝑢𝑥1 =
= 𝑥1
𝜕𝑥1
𝜕𝑥2
𝑀𝑅𝑆 =
𝑋2 + 2 4
= =2
𝑋1
2
→ 2𝑋1 = 𝑋2 + 2 → 𝑋1 =
𝑋2 + 2
𝑋1
103
𝑠𝑢𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑏𝑢𝑑𝑔𝑒𝑡 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 60=4𝑥1 +2𝑋2→ 60=4 (0.5 𝑋2+1) +2𝑋2
=4𝑋2 +4 → 𝑋2=56 𝑋2 =14, 𝑋1=8
The sufficient condition is satisfied since the indifference curve is convex to
the origin. you could prove that MRS is decreasing as 𝑥1increases
𝜕𝑀𝑅𝑆
𝜕𝑥1
< 0.
3) Lagrange Multiplier Method
•First order condition
•Second order condition.
For a constrained optimum of Z =f (X, Y) subject to constraint C=g (X, Y). We
writedown Lagrange function that consist of Two conditions must be satisfied
for optimum.
The necessary or first order condition which consist of the first derivatives
of (Z)
with respect to (X, Y, λ) equating each one to zero. ( 𝑍𝑋,𝑍𝑌,𝑍𝜆 =0).
The sufficient or second order condition can be expressed in the Bordered
104
Hessian
⌈𝐻⌉. Form, which consist of the second partial derivatives of the necessary
conditions as follow:
𝑍𝑋𝑋
| 𝑍𝑌𝑋
𝑍𝜆𝑋
If f < 0→
𝑍𝑋𝑌
𝑍𝑌𝑌
𝑍𝜆𝑌
𝑍𝑋𝜆
𝑍𝑌𝜆 | If f > 0 → optimum maximum.
𝑍𝜆𝜆
optimum minimum
Let us return to the previous example to check the sufficient condition. We
startwith the first order condition.
Z= 𝑋1𝑋2 + 2𝑋1 + λ (60 -4𝑋1-2𝑋2)
𝑍𝑋1 =𝑋2+2 - 4λ = 0 →
𝑋2+2= 4λ
(1)
𝑍𝑋2 = 𝑋1-2λ = 0→ 𝑋1 =2λ
(2)
𝑍𝜆= 60 -4𝑋1-2𝑋2 = 0 →
(3)
Dividing (1) on (2):
𝑋2 + 2 4𝜆
=
= 2 → 2𝑥1 = 𝑥2+2 → 𝑥1 = 0.5𝑥2 + 1
𝑋1
2𝜆
substitute in budget constraint, 60 - 4 (0.5𝑥2 +1)-2𝑥2
=0
60 − 2𝑥2-4=2𝑥2 →4𝑥2=56 → 𝑥2=14, 𝑥1=0.5(14) +1=8, 4
Sufficient condition:
0
1 −4
1
|𝐻| = | 1
0 −2| = −1 |
−4
−4 −2 0
−2
1
0
| −4|
|
0
−4 −2
−1(0 − 8) − 4(−2 − (−4) (0) = 8 + 8 = 16 > 0 (𝑀𝑎𝑥𝑖𝑚𝑢𝑚)
105
𝑥1∗ = 8, 𝑥2∗ = 14, 𝜆 = 4
4) Some Economic applications
a) Utility Maximization.
Suppose the consumer spend all his income only on two goods (X and Y),
both have continuous positively marginal utility functions, the prices of
both goods are given determined by market (,), consumer’s income is
given at (I).if utility function is U=f (X,Y), (𝑼𝑿 ,𝑼𝒀) > 𝟎
Subject to budget constraint: P. 𝑿 +𝑷𝒀. 𝒀 = I
First order condition:
Set the first partial derivatives of Z with respect to (X, Y, λ) equal to zero.
The Lagrange function of utility maximization.
Z = U (X, Y) + λ (I - Px.X -𝑷𝒀. 𝒀 ).
𝒁𝑿=𝑼𝑿 - λ𝑷𝑿= 0 → 𝑼𝑿=𝛌 𝑷𝑿
(1)
𝒁𝒀=𝑼𝒀 - λ𝑷𝒀 = 0 → 𝑼𝒀 = λ𝑷𝒀→
(2)
𝒁𝝀=( I – P x . X - -𝑷𝒀. 𝒀) = 0
(3)
𝑼𝑿
Dividing (1) on (2) we get:
𝑼𝒀
=
𝑷𝑿
𝑷𝒀
𝒐𝒓
𝑼𝑿
𝑷𝒙
=
𝑼𝒀
𝑷𝒀
=𝝀
(4)
Equation (4) indicates that the consumer must allocate his budget to
106
equalize the ratio of marginal utility to price of each good. this give the
marginal utility of money when the consumer is maximizing utility (λ).
Second order condition:
If the bordered Hessian (H) is positive utility is maximum.
𝑍𝑋𝑋
|𝐻| = | 𝑍𝑌𝑋
𝑍𝜆𝑋
𝑍𝑋𝑌
𝑍𝑌𝑌
𝑍𝜆𝑌
𝑍𝑋𝜆
𝑢𝑋𝑋
𝑍𝑌𝜆 | = |𝑍𝑢𝑌𝑋
𝑍𝜆𝜆
−𝑃𝑋
𝑢𝑋𝑌
𝑢𝑌𝑌
−𝑃𝑌
𝑃𝑋
−𝑃𝑦 | > 0
0
b)Constrained output maximization:
Suppose the firm seek to maximize output Y=f(x1, 𝒙𝟐) subject to a cost
constraint 𝑪𝟎= 𝒓𝟏𝒙𝟏 +𝒓𝟐𝒙𝟐 .the firm is seeking to obtain the greatest
possible output for a given cost level. The problem is to maximize output Y
=f(x1, 𝒙𝟐) subject to 𝑪𝟎= 𝒓𝟏𝒙𝟏 +𝒓.
Z= f (x1, x2) + 𝝀 (𝑪𝟎- 𝒓𝟏𝒙𝟏-𝒓𝟐𝒙𝟐)
The first order condition: the first partial derivatives of Z with respect to
(X1,, X2 , λ).is zero.
𝒁𝑿𝟏 =𝒇𝟏 – 𝝀 𝒓𝟏
=0
→ 𝒇𝟏 = 𝝀 𝒓𝟏
(1)
𝒁𝑿𝟐 =𝒇𝟐 - λ𝒓𝟐 =0
→
𝒇𝟐 = λ𝒓𝟐
(2)
𝑍𝜆 = (𝐶0 − 𝑟1𝑥1 − 𝑟2𝑥2 ) = 0 → 𝑋2 = 𝑋2 =
Dividing (1) on (2) we get:
𝑓1
𝑓2
𝑟
𝑓1
𝑟2
𝑓1
− ( 1 ) 𝑜𝑟
=
𝑓2
𝑓2
𝐶0
𝑟1
− ( ) 𝑋1
𝑟2
𝑟2
(3)
=𝜆
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This means the contribution to output of spending last dollar on each
inputmust be equal to λ, Notice MRTS can be calculated as
𝒇
𝑴𝑹𝑻𝑺 = 𝟏 =.𝒓𝟐
𝒇𝟐
Second order condition:
𝑓11
|𝐻| => 0 | 𝑓21
−𝑟1
𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐬 𝐭𝐡𝐚𝐭
𝑓12
𝑓22
−𝑟2
−𝑟1
−𝑟2 | > 0
0
This can be shown graphically by the tangency point between the iso-cost
𝒓
𝒅𝒙
line(the slope) = − ( 𝟏 )and the iso-quant curve) the slope = ( 𝟐 ) →
𝒓𝟏
𝒓𝟐
𝒅𝒙𝟏
𝑴𝑹𝑻𝑺 = ( ). This is the first order condition; the second order condition is
𝒓𝟐
satisfied when the iso-quant is convex to the origin (MRTS) is decreasing.
❖ Example: Given the production function is Q =f (X, Y) =X (Y+ 4) ,
TC is given by: X + Y =8.
*Find the maximum output subject to cost constraint, and check the second
order condition?
Z = X (Y + 4) + (8-X- Y)
Z = (XY+ 4X) + (8-X- Y)
First order condition:
𝒁𝑿=(𝒀+𝟒) − 𝝀 = 0 → Y+4 =λ
(1)
𝒁𝒀=X -λ =0 →
(2)
X =λ
𝒁𝝀= (8-X- Y) =0 →
X= 8- Y
Dividing (1) on (2) we get:
𝒀+𝟒
𝑿
(3)
=𝟏 → 𝐗 =𝐘 +𝟒
By substituting in (3): 8- Y=Y +4 → 𝒀 = 𝟐
X = Y +4= 6, λ =6, Q =6 (2+4) =36
Second order condition:
108
0
1 −1
1
|𝐻| = | 1
0 −1| = −1 |
−1
−1 −1 0
−1
1
0
| −1|
|
0
−1 −1
|𝑯|= -1(0 -1) -1 (-1+0) = 2 > 𝟎 → (Maximum)
c)Cost minimization:
The firm may desire to minimize the cost of producing a given level of
output.
The problem is to minimize the objective function (iso-cost function
TC=𝑟1𝑥1+𝑟2𝑥2
subject to a constraint (a given output Y̅ ) .Y = f (x1 x2,) = Y̅
The Lagrange function is given by:
Z =𝒓𝟏𝒙𝟏 +𝒓𝟐𝒙𝟐 + λ (Y̅ - f (x1, x𝟐)
The first order condition is to set the first partial derivatives of Z with respect
to (X, Y, λ) equal to zero.
𝒁𝑿𝟏 =𝒓𝟏 - λ𝒇𝟏 = 0→ 𝒓𝟏 = λ𝒇𝟏
(1)
𝒁𝑿𝟐 =𝒓𝟐−𝝀𝒇𝟐 = 0 →
𝒓𝟐=𝝀𝒇𝟐
(2)
𝒁𝝀= (Y̅ - f ( , 𝑿𝟐) →
Y̅ =- f (𝑿𝟏 , 𝑿𝟐)
(3)
Dividing equation (1) on (2) we get:
𝑓1
𝑓2
𝑟
𝑓1
𝑟2
𝑟
− ( 1 ) = 𝑀𝑅𝑇𝑆 𝑜𝑟
=
𝑓2
𝑟2
1
= , is the marginal cost of production at
𝜆
optimum.
109
The second order condition:
𝜆𝑓11
|𝐻| => 0 |𝜆𝑓21
−𝑓1
𝜆𝑓12
𝜆𝑓22
−𝑓2
−𝑓1
−𝑓2 | > 0 (𝑚𝑖𝑛𝑖𝑚𝑢𝑚)
0
This can be shown graphically as follows. The least cost is TC1.The optimal
input choice (𝒍∗, k∗) to produce q1.
❖ Example: A producer desires to minimize his cost of production
C = 2L+5Ksubject to the production constraint Q =L.K.
*Find the optimum combination of L, K to minimize cost of producing 40
units of output.? Calculate the least cost.
Z =2L+5K + λ (40 – LK)
(Lagrange function)
The first order condition for minimization requires:
𝒁𝑳 =2- λ K = 0
→2= λ K
(1)
𝒁𝑲=5 – λL = 0
→ 5 =λL
(2)
𝒁𝝀= 40 -LK =0
→ 40=LK
(3)
Dividing (1) on (2) we get:
𝐾
𝐿
2
= →
5
2𝐿 = 5𝐾 → 𝐿 =
5
𝐾
2
→
substituting in (3)
40 =
5
𝐾. 𝐾 → 80 = 5𝐾 2 = → 𝐾 2 = 16 →
2
5
2 5
𝐾 = 4, 𝐿 = . 4 = 10 , 𝜆 = =
2
𝐾 𝐿
=0.5
110
Second order condition for minimum requires H< 𝟎
0
|𝐻| = | −𝜆
−𝐾
−𝜆
0
−𝐿
|𝐻| = −(−0.5) |
−𝐾
0
|
=
|
−𝐿
0.5
0
−4
−0.5 −4
0
−10|
−10
0
−0.5 −10
−0.5
| − 4|
−4
0
−4
0
|
10
|𝑯|=0.5(-40) -4(5) =- 40 < 𝟎
Least cost of production= 2(10) +5(4) =40
Assignment:
1-Given U = (X +2) (Y+ 1) and =4, P y =6 and I =130
*Write the Lagrange function.
*Find the optimal values of X and Y. Calculate 𝜆 and U.
*Is the sufficient condition is satisfied for maximum.
111
2-Find the Optimum of Y =𝑋2 + 𝑋2 Subject to 𝑥1 +4𝑋2 =2. Use Lagrange
Multiplier Method and check the sufficient condition.
3-Use Lagrange Multiplier to find the maximum or minimum of the following
functions:
•
Z = X. Y Subject to X +Y = 6.
•
Z =X -3Y-X. Y Subject to X +Y =6
4-Suppose the production function is Given by:
𝑄 = 𝑓(𝑥1, 𝑥2 ) = 4𝑋10.5𝑋20.5 = 𝑄
Suppose the cost function is TC =2𝑋1 +8𝑋2
•
Find the optimal choice of (𝑥1, 𝑥2) that minimize cost if production equal
(Q̅) =32. assuming the sufficient condition exists.
5-Suppose the utility function is Given by: U (𝑥1, 𝑥2)
=2𝑋1 .𝑋2
Suppose the consumer’s income I and prices of the two goods are𝑃𝑋1and 𝑝𝑥2.
•
Find the demand functions of 𝑥1 𝑎𝑛𝑑 𝑥2.using Lagrange Multiplier.
•
Determine the degree of homogeneity of these functions. And explain
what does it mean.
112
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