Managerial Economics
in a Global Economy
Chapter 2
Optimization Techniques
and New Management Tools
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
This form can be expressed in an equation:
 Q=f(P)
(1)
 Though useful, it does not tell us how Q responds to P, but
this equation do.
 Q = 200 - 5 p
(2)
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
Note: Total profit is maximized when marginal profit shifts from positive to negative.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Average Profit = Profit / Q
 Slope of ray from the
origin:
PROFITS
MAX
 Rise / Run
 Profit / Q = average profit
C
B
 Maximizing average profit
doesn’t maximize total
profit
profits
Q
Managerial Economics
quantity
Prof. M. El-Sakka
CBA. Kuwait University
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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  B
2005 South-Western
Publishing
Managerial
Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
the rise (Y) over the
run (X)
 The derivative at point
C is also this slope
D
Y
Y
X
C
X
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Optimum Can Be Highest or Lowest
 Finding the maximum flying range for the Stealth
Bomber is an optimization problem.
 Calculus teaches that when the first derivative is zero, the
solution is at an optimum.
 The original Stealth Bomber study showed that a
controversial flying V-wing design optimized the
bomber's range, but the original researchers failed to find
that their solution in fact minimized the range.
 It is critical that managers make decision that maximize,
not minimize, profit potential!
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Quick Differentiation Review
Name
Function
Derivative
Example
 Constant Y = c
Functions
dY/dX = 0
Y=5
dY/dX = 0
Y = c • X dY/dX = c
Y = 5X
dY/dX = 5
 A Line
 Power Y = cXb
Functions
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dY/dX = b•c•X b-1 Y = 5X2
dY/dX = 10X
Prof. M. El-Sakka
CBA. Kuwait University
Quick Differentiation Review
 Sum of Y = G(X) + H(X)
Functions
example Y = 5X + 5X2
dY/dX = dG/dX + dH/dX
dY/dX = 5 + 10X
 Product of Y = G(X) • H(X)

Two Functions
dY/dX = (dG/dX)H + (dH/dX)G
example
Y = (5X)(5X2 )
dY/dX = 5(5X2 ) + (10X)(5X) = 75X2
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
 USING DERIVATIVES TO SOLVE MAXIMIZATION AND
MINIMIZATION PROBLEMS
 Maximum or minimum points occur only if the slope of the
curve equals zero.
 Look at the following graph
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
0
dY = 100 - 10X
dX
dY = 0
if
dX
X = 10
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
y
Max value of y
Since
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
value of dy/dx
d2y/dx2 > 0
d2y/dx2 < 0
d 2Y = > 0
dX 2
( minimum point )
* at point 10 the curve is downward, i.e., its slope is
negative. Hence
d 2Y = < 0
dX 2
( maximum point )
x
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
 Graphs of an original third-order function and its first and second
derivatives. (what if the second order = 0)
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Optimization Rules
Maximization conditions:
1-
2-
dY = 0
dX
d 2Y = < 0
dX 2
Minimization conditions:
1-
2Managerial Economics
dY = 0
dX
d 2Y = > 0
dX 2
Prof. M. El-Sakka
CBA. Kuwait University
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
More Examples
 Competitive Firm: Maximize Profits
 where  = TR - TC = P • Q - TC(Q)
 Use our first order condition:
d/dQ = P - dTC/dQ = 0.
 Decision Rule: P = MC.

a function of Q
Max = 100Q - Q2
First order = 100 -2Q = 0 implies
Q = 50 and;
 = 2,500
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Second Order Condition: one variable
 If the second derivative is negative, then it’s a maximum
 If the second derivative is positive, then it’s a minimum
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Extra examples
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
d 2Y
dX 2
= -12 + 6X
at X = 3
d 2Y
dX 2
at X = 1
b = -12
d 2Y
dX 2
Managerial Economics
or X = 1
the second condition
a=3
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)
Prof. M. El-Sakka
CBA. Kuwait University
Partial Differentiation
 Economic relationships usually involve several
independent variables.
 A partial derivative is like a controlled experimentit 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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Problem:
 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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Solution: 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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
partial
(2)
derivatives
equal
to
zero
and
solve
simultaneously
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
step 1:
30 - Q1 - Q2 = 0
step 2:
L = 4Q 12 + 5Q22 - Q1Q2 + ( 30 - Q1 - Q2)
step 3:
L
Q1
= 8Q 1 - Q2 - 
L
Q2
= -Q1 + 10Q 2 - 
L

= -Q1 - Q2 + 30
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
8Q1 - Q2 -  = 0
(1)
-Q1 + 10Q2 -  =0
(2)
-Q1 - Q2 + 30 =0
(3)
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
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
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.
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University
Key Terms













Marginal profit
Average profit
Marginal cost
Marginal revenue
Marginal analysis
Optimization
Derivative of Y with respect to X (dy/dx)
Differentiation
Second derivative
Partial derivative
Constrained optimization
Lagrangian multiplier method
Lagrangian function
Managerial Economics
Prof. M. El-Sakka
CBA. Kuwait University