Managerial Economics
Dr Nihal Hennayake
What is Microeconomics and Macroeconomics ?
•Ragnor Frisch : Micro means “ Small” and Macro
means “Large”
Microeconomics deals with the study of individual
behaviour.
• It deals with the equilibrium of an individual consumer,
producer, firm or industry.
Macroeconomics on the other hand, deals with economy
wide aggregates.
• Determination of National Income Output, Employment
• Changes in Aggregate economic activity, known as Business
Cycles
• Changes in general price level , known as inflation, deflation
• Policy measures to correct disequilibrium in the economy,
Monetary policy and Fiscal policy
What is Managerial Economics?
“Managerial Economics is economics applied in decision making.
It is a special branch of economics bridging the gap between
abstract theory
and managerial practice” – Willian
Warren Haynes, V.L. Mote, Samuel Paul
“Integration of economic theory with business practice for the
purpose of facilitating decision-making and forward planning” Milton H. Spencer
“Managerial economics is the study of the allocation of scarce
resources available to a firm or other unit of management among
the activities of that unit”
Willian Warren
Haynes, V.L. Mote, Samuel Paul
“ Price theory in the service of business executives is known as
BUSINESS ADMINISTRATION
DECISION PROBLEMS
TRADITIONAL ECONOMICS :
THEORY AND METHODOLOGY
DECISION SCIENCES :
TOOLS AND TECHNICS
MANAGERIAL ECONOMICS :
INTEGRATION OF ECONOMIC
THEORY AND
METHODOLOGY WITH TOOLS
AND TECHNICS BORROWED
FROM OTHER DECIPLINES
OPTIMAL SOLUTIONS TO
BUSINESS PROBLEMS
Nature, Scope and Significance of Managerial Economics:
 Managerial Economics – Business Economics
 Managerial Economics is ‘Pragmatic’
 Managerial Economics is ‘Normative’
 Universal applicability
 The roots of Managerial Economics came from Micro
Economics
 Relation of Managerial Economics to Economic Theory is much
like that of Engineering to Physics or Medicine to Biology. It is the
relation of applied field to basic fundamental discipline
Core content of Managerial Economics :

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Demand Analysis and forecasting of demand
Production decisions (Input-Output Decisions)
Cost Analysis (Output - Cost relations)
Price – Output Decisions
Profit Analysis
Investment Decisions
1. Demand Analysis :
Meaning of demand : No. of units of a commodity that
customers are willing to buy at a given price under a set of
conditions.
Demand function : Qd = f (P, Y, Pr W)
Demand Schedule : A list of prices and quantitives and
the list is so
arranged that at each price the
corresponding
amount is the quantity purchased at
that price
Demand curve
: Slops down words from left to right.
Law of demand
: inverse relation between price and
quantity
Exceptions to the law of demand :
Giffens paradox
Price expectations
Elasticity : Measure of responsiveness - Qd = f (P, Y, Pr W)
E = percentage change in DV/ percentage change in IV
Concepts of price, income, and cross elasticity
Price Elasticity :
Ep = Percentage change in QD/Percentage change in P
Types of price elasticity :
1. Perfectly elastic demand Ep = ∞
2. Elastic demand Ep > 1
3. Inelastic demand Ep < 1
4. Unit elastic demand Ep = 1
5. Perfectly inelastic demand Ep = 0

Elasticity and expenditure : If demand is elastic a
given fall in price causes a relatively larger increase in
the total expenditure.
P↓ - TR↑ when demand is elastic.
 P↓ - TR↓ when demand is inelastic.
 P↓ ↑ - TR remains same when demand is Unit elastic.
Measurement of elasticity :
 Point and Arc elasticity
 Elasticity when demand is linear
 Determinants of elasticity :
(1) Number and closeness of its substitutes,
(2) the commodity’s importance in buyers’ budgets,
(3) the number of its uses.
 Other Elasticity Concepts
 Income elasticity
 Cross elasticity
Functions of a Managerial Economists:
The main function of a manager is decision making and
managerial
Economics helps in taking rational decisions.
The need for decision making arises only when there are more
alternatives courses of action.
Steps in decision making :
Defining the problem
Identifying alternative courses of action
Collection of data and analyzing the data
Evaluation of alternatives
Selecting the best alternative
Implementing the decision
Follow up of the action

Specific functions to be performed by a managerial Economist :
1. Production scheduling
2. Sales forecasting
3. Market research
4. Economic analysis of competing companies
5. Pricing problems of industry
6. Investment appraisal
7. Security analysis
8. Advice on foreign exchange management
9. Advice on trade
10.Environmental forecasting
OPTIMIZATION
Managerial economics is concerned with the ways in
which managers should make decisions in order to
maximize the effectiveness or performance of the
organizations they manage. To understand how this can
be done we must understand the basic optimization
techniques.
Functional relationships:
relationships can be expressed by graphs:
P
Q
This form can be expressed in an equation:
Q=f(P)
Though useful, it does not tell us how Q responds to P,
but this equation do.
Q = 200 - 5 p
Marginal Analysis
The marginal value of a dependent variable is
defined as the change in this dependent
variable associated with a 1-unit change in a
particular independent variable. e.g.
units

0
0
marginal profit
average profit
-
100
1
100
100
150
2
250
125
350
3
600
200
400
4
1000
250
350
5
1350
270
150
6
1500
250
50
7
1550
221
-50
8
1500
188
-100
9
1400
156
Total profit is maximized when marginal profit shifts from positive to
negative.
Average Profit = Profit / Q
PROFITS
Slope of ray from the
origin:
MAX
C
B
Maximizing average
profit doesn’t
maximize total profit
profits
Q
– Profit / Q = average
profit
quantity
Marginal Profits = /Q
 Q1 is breakeven (zero
profit)
 maximum marginal
profits occur at the
inflection point (Q2)
 Max average profit at
Q3
 Max total profit at Q4
where marginal
profit is zero
 So the best place to
produce is where
marginal profits = 0.
profits
(Figure 2.1)
max
Q3
Q4
Q2
Q1
Q
average
profits
marginal
profits
Q
Differential Calculus in
Management
A function with one decision variable, X, can be written
as:
Y = f(X)
The marginal value of Y, with a small increase of X, is
My = Y/X
For a very small change in X, the derivative is written:
dY/dX = limit Y/X
X  0
Marginal = Slope = Derivative
The slope of line C-D
is Y/X
The marginal at point
C is Y/X
The slope at point C
is Y over X
The derivative at
point C is also this
slope
D
Y
Y
X
C
X
Quick Differentiation Review
Name
Function
Constant Y = c
Functions
A Line
Derivative
Example
dY/dX = 0
Y=5
dY/dX = 0
Y = c • X dY/dX = c
Power Y = cXb
Functions
Y = 5X
dY/dX = 5
dY/dX = b•c•X b-1 Y = 5X2
dY/dX = 10X
Quick Differentiation Review
Sum of Y = G(X) + H(X)
Functions
example Y = 5X + 5X2
Product of
dY/dX = dG/dX + dH/dX
dY/dX = 5 + 10X
Y = G(X) • H(X)
Two Function dY/dX = (dG/dX)H + (dH/dX)G
example
Y = (5X)(5X2 )
dY/dX = 5(5X2 ) + (10X)(5X) = 75X2
Quick Differentiation Review
Quotient of Two Y = G(X) / H(X)
Functions
dY/dX = (dG/dX)•H - (dH/dX)•G
H2
Y = (5X) / (5X2) dY/dX = 5(5X2) -(10X)(5X)
(5X2)2
= -25X2 / 25X4 = - X-2
Chain Rule
Y = G [ H(X) ]
dY/dX = (dG/dH)•(dH/dX) Y = (5 + 5X)2
dY/dX = 2(5 + 5X)1(5) = 50 + 50X
Max of x
y
Slope = 0
the function
Y = -50 + 100X - 5X2
value of x
i.e.,
dy/dx
10
20
x
Value of Dy/dx when y is max
dY = 100 - 10X
dX
dY = 0
if
dX
X = 10
0
Value of dy/dx which
Is the slope of y curve
i.e., Y is maximized when
the slope equals zero.
10
20
x
Note that this is not sufficient for maximization or minimization problems.
Since
y
Max value of y
dY
dX
= 0 at two points, we need another
condition to distinguish between the maximum and
minimum points.
Look at the
Min value of y
dY
dX
curve
* at point 5 the curve is upward, i.e., its slope ( the
second derivative (the derivative of the derivative)) is
positive. Hence
x
Dy/dx
d 2Y = > 0
dX 2
value of dy/dx
d2y/dx2 > 0
( minimum point )
d2y/dx2 < 0
* at point 10 the curve is downward, i.e., its slope is
negative. Hence
d 2Y = < 0
dX 2
x
( maximum point )
Optimization Rules
Maximization conditions:
1-
2-
dY = 0
dX
d 2Y = < 0
dX 2
Minimization conditions:
1-
2-
dY = 0
dX
d 2Y = > 0
dX 2
Applications of Calculus in Managerial
Economics
maximization problem:
A profit function might look like an arch, rising to a peak and then
declining at even larger outputs. A firm might sell huge amounts
at very low prices, but discover that profits are low or negative.
At the maximum, the slope of the profit function is zero. The first
order condition for a maximum is that the derivative at that point
is zero.
If  = 50Q - Q2,
then d/dQ = 50 - 2·Q, using the rules of differentiation.
Hence, Q = 25 will maximize profits
where
50 - 2Q = 0.
More Applications of Calculus
minimization problem: Cost minimization
supposes that there is a least cost point to produce. An
average cost curve might have a U-shape. At the
least cost point, the slope of the cost function is zero.
The first order condition for a minimum is that the
derivative at that point is zero.
If TC = 5Q2 – 60Q,
then dC/dQ = 10Q - 60.
Hence, Q = 6 will minimize cost
Where:
10Q - 60 = 0.
Competitive Firm: Maximize Profits
– where  = TR - TC = P • Q - TC(Q)
– Use our first order condition:
– d/dQ = P - dTC/dQ = 0.
a function of Q
– Decision Rule: P = MC.
Max = 100Q - Q2
First order = 100 -2Q = 0 implies
Q = 50 and;
 = 2,500
Second Order Condition: one variable
If the second derivative is negative,
then it’s a maximum
.
Problem 1
Problem 2
Max = 100Q - Q2
Max= 50 + 5X2
First derivative
First derivative
100 -2Q = 0
second derivative is: -2
implies
Q =50 is a MAX
10X = 0
second derivative is: 10
implies
Q = 10 is a MIN
e.g.;
2
3
Y = -1 + 9X - 6X + X
Y = 0 at
2
= 9 - 12X + 3X = 0
Quadratic Function
2
Y = aX + bX + c
X=
b  b2  4ac
2a
X=3
or X = 1
the second condition
d 2Y
dX 2
= -12 + 6X
at X = 3
d 2Y
dX 2
a=3
at X = 1
b = -12
d 2Y
dX 2
c=9
=2  1
therefore
first condition
dY
dX
X=
(12)  122  4(9  3)
6
= -12 + 6(3) = 6 >0 ( minimum point)
= -12 + 6(1) = - 6 <0 (maximum point)
Partial Differentiation
Economic relationships usually involve
several independent variables.
A partial derivative is like a controlled
experiment- it holds the “other” variables
constant
Suppose price is increased, holding the
disposable income of the economy constant
as in
Q = f (P, I )
then Q/P holds income constant.
Sales are a function of advertising in
newspapers and magazines ( X, Y)
Max S = 200X + 100Y -10X2 -20Y2 +20XY
Differentiate with respect to X and Y and set
equal to zero.
S/X = 200 - 20X + 20Y= 0
S/Y = 100 - 40Y + 20X = 0
solve for X & Y and Sales
2 equations & 2 unknowns
200 - 20X + 20Y= 0
100 - 40Y + 20X = 0
Adding them, the -20X and +20X cancel, so
we get 300 - 20Y = 0, or Y =15
Plug into one of them:
200 - 20X + 300 = 0, hence X = 25
To find Sales, plug into equation:
S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250
PARTIAL DIFFERENTIATION AND MAXIMIZATION OF
MULTIVARIATE FUNCTIONS.

= f (Q1 , Q2 )
To know the marginal effect of Q1 on

we hold Q2 constant, and
vice versa.
In order to do that we use partial derivative of
Q1 denoted by

with respect to
 ( treating Q2 as constant )
Q1
e.g.;

= -20 + 100Q1 + 80Q2 - 10Q12 - 10Q22 - 5Q1Q2;
to find the partial derivative of

with respect to Q1 we treat Q2
as constant; hence

Q1
= 100 - 20Q1 - 5Q2;
(1)
therefore

Q2
= 80 - 20Q2 - 5Q1;
setting
both
simultaneously
partial
(2)
derivatives
equal
to
zero
and
solve
100 - 20Q1 - 5Q2 =0
80 - 20Q2 - 5Q1 =0
multiply by -4 and add
________________
- 220 + 75Q2 = 0
hence
Q2 = 2.933
substitute for Q2 at any of the eq. 1
100 - 20Q1 - 14.665;
hence
Q1 = 4.267.
i.e.,
profit is maximized when the firm produces 4.267 of Q1 and 2.933 of Q2.
CONSTRAINED OPTIMIZATION
We assume that the firm can freely produce 4.267 of Q1 and 2.933
of Q2. Quite often this may not be the case.
e.g.
Minimize TC = 4Q12 + 5Q22 - Q1Q2;
subject to:
Q1 + Q2 = 30
The constraint function
Solution:
The lagrangian multiplier:
Steps:
1 - set the constraint function to zero
2 - form the lagrangian function by adding the constraint function
after multiplication with an unknown factor  to the original
function.
3 - take the partial derivatives and set them equal to zero
4 - solve the resulting equations simultaneously
step 1:
30 - Q1 - Q2 = 0
step 2:
L = 4Q 12 + 5Q 22 - Q1Q2 + ( 30 - Q1 - Q2)
step 3:
L
Q1
L
Q 2
L

= 8Q 1 - Q2 - 
= -Q1 + 10Q 2 - 
= -Q1 - Q2 + 30
8Q1 - Q2 -  = 0
(1)
-Q1 + 10Q2 -  =0
(2)
-Q1 - Q2 + 30
(3)
=0
step 4
multiply eq(2) by -1 and subtract from eq(1)
9Q1 - 11Q2 = 0
(4)
multiply (3) by 9 and add to eq(4)
-9Q1 - 9Q2 + 270 = 0
9Q1 - 11Q2
=0
____________________
-20Q2 +270 = 0
Q2 = 270/20 = 13.5
substituting in eq (3) Q1 = 16.5
the values of Q1 and Q2 that minimizes TC are 16.5 and 13.5
respectively.
substituting Q1 and Q2 in eq(1) or eq(2) we find that
 = 118.5
the interpretation of 
measures the change in TC if the constraint is to be relaxed by one
unit.
i.e., TC will increase ( has a positive sign ) by 118.5 if the constraint
becomes 29 or 31.