Transcript Price Elasticity of Demand

Lecture 2
Derivatives
dY
Y
 lim
dX X 0 X
Derivative of a constant
Y
dY
Y
Y 1 Y 1
 lim
 lim

X

0
dX
X X 0 X 2  X 1
Y=3
Y1
dY
0
 lim
0
dX X 0 X 2  X 1
X1
X2
X
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Lecture 2
Derivative of a line
dY
Y
Y 2 Y 1
 lim
 lim
dX X 0 X X 0 X 2  X 1
Y
dY
15  10 5
 lim
 5

X

0
dX
32
1
Y=5X
Y2=15
Y1=10
X1=2 X2=3
X
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Lecture 2
Derivative of a polynomial function
Y  ab X n  ab 1 X ( n 1 )  ab  2 X ( n  2 )  ......  a0
dY
 nab X ( n 1 )  ( n  1 )ab 1 X ( n  2 )  ( n  2 )ab  2 X ( n 3 )  ......  a1
dX
Examples:
Y  3X
dY
 3  2  X 2 1  6 X
dX
2
Y  KX  CX  3 X  1
3
2
dY
 3KX 2  2CX  3
dX
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Lecture 2
Derivatives of sums and differences
In general:
For
Y ( x )  W ( x )  Z( x )
dY dW dZ


dX dX dX
or
Y ( x )  W ( x )  Z( x )
dY dW dZ


dX dX dX
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Lecture 2
Derivatives of products and quotients
In general:
Y ( x )  W ( x )Z( x )
For
dY
dZ
dW
W
Z
dX
dX
dX
Examples:
Y( x )  6 X( 3  X 2 )
W( x )  6X
Z( x )  3  X
dY
dZ
dW
6X
(3 X 2 )
dX
dX
dX
dY
 6 X ( 2 X )  ( 3  X 2 )6
dX
 12 X 2  18  6 X 2
2
 18  18 X  18( 1  X )
2
2
or
Y ( x )  W ( x ) / Z( x )
dY Z(

dX
dW
dX
5X 3
Y( x ) 
34X
W( x )  5X 3
)  W ( dZ
dX
Z2
Z( x )  3  4 X
dW
 15 X 2
dX
dZ
 4
dX
dY ( 3  4 X )( 15 X 2 )  5 X 3 ( 4 )

dX
( 3  4 X )2
45 X 2  40 X 3 X 2 ( 45  40 X )


( 3  4 X )2
( 3  4 X )2
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)
Lecture 2
Derivative of a derivative
Profit
d 2Y
d dY

(
)
2
dX
dx dX
A
Y  4 X 2  7 X  3
0
Q
1
Number of
units of output
dY
 8 X  7
dX
d 2Y
 8
dX 2
Y  3X 2  4 X 1
Profit
dY
6X 4
dX
d 2Y
6
dX 2
A
0
Q
1
Number of
units of output
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Lecture 2
Partial Derivative
A derivate with respect to only one variable when the function is the
function of more than just that variable
A single variable function:
Y( x )  3X 3  4 X  1
dY
 9X 2  4
dX
A multi-variable function:
Y ( x ,w )  3 X 3  4 XW 2  2 X 2  W  1
Y
 9 X 2  4W 2  4 X
X
Y
 0  8 XW  0  1  8 XW  1
W
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Lecture 2
Optimization Theory
Unconstrained Optimization
Unconstrained optimization applies when we wish to find the maximum or
minimum point of a curve. In other words we wish to find the value of the
independent variable at which the dependent variable is maximized or
minimized without any other external conditions restricting it.
Let us assume that there is an activity x which generates both
value V(x) and cost C(x).
Net value would therefore be:
The necessary condition to find the
optimal level is:
Or:
NV ( x )  V ( x )  C( x )
dNV ( x )
dV ( x ) dC( x )
0 

dx
dx
dx
dV ( x ) dC( x )

dx
dx
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Lecture 2
Unconstrained Optimization: Multiple variables
In the case where there are more than one activity, say when the value
function is a function of x and y, we take the derivative of the function with
respect to each variable and set them all to zero.
NV ( x , y )  V ( x , y )  C( x , y )
NV ( x , y ) V ( x , y ) C( x , y )


0
x
x
x
NV ( x , y ) V ( x , y ) C( x , y )


0
y
y
y
V ( x , y ) C( x , y )

x
x
and
V ( x , y ) C( x , y )

y
y
As such we have:
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Lecture 2
Example:
ABCO LLC has two product lines: gadgets and widgets.
ABCO produces G of gadgets and W of widgets annually
The profit made by ABCO is of course related to their quantity of
widgets and gadgets sold. The following equation shows this
relationship:
P( G ,W )  5GW  10G 2  10W 2  80W  113.75G  20
Find the derivative (partial derivative) of profit with respect to G.
P
 5W  20G  113.75
G
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Lecture 2
Find the derivative (partial derivative) of profit with respect to W.
P
 5G  20W  80
W
Now, using this information find the quantities of G and W that ABCO
must manufacture to maximize profit.
To answer this question, we remember that a point is either a maximum or
minimum when the derivative for that point is zero. For P to be maximized
both derivatives with respect to G and W must be zero.
 5W  20G  113.75  0
 5G  20W  80  0
G  5.0
W  2.75
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Lecture 2
Constrained Optimization
Constrained optimization applies when we wish to find the maximum or
minimum point of a curve but there are also other limiting factors. In other
words we wish to find the value of the independent variable at which the
dependent variable is maximized or minimized with other external conditions
restricting it.
Let us start – without loss of generality -with
the marginal value for a two variable case:
The constraint is that the total cost must
equal a specified level of cost relating to
the price and quantities of the two
components x, and y:
V ( x , y )
V ( x , y )
and
x
y
C( x , y )  Px x  Py y  C
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Lecture 2
There are two equivalent ways of solving such problems:
1. Simple simultaneous equations:
In this approach we solve the set of equations:
V ( x , y )
0
x
V ( x , y )
0
y
Px x  Py y  C  0
2. Lagrangian method:
The Lagrangian method works on the basis of adding “meaningful zeros”
to the original equation and then assess their impact.
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Lecture 2
The first thing we do is to form the Lagrangian function.
To do so, we first rearrange our constraint formula or formulas so that they
all evaluate to zero:
Px x  Py y  C
Px x  Py y  C  0
Then, we add “zero” to the original value function:
L  V ( x , y )  ( C  Px x  Py y )
Now we take partial derivatives of the value function wrt x, y and λ, set
these to zero and solve.
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Lecture 2
Example:
Cando Co wishes to minimize the cost of their production governed by:
TC  4Q12  5Q22  Q1Q2
The constraint is that the company can only make 30 units of product in total
Q1  Q2  30 or
The Lagrangian becomes:
As such we have:
30  Q1  Q2  0
L  4Q12  5Q22  Q1Q2  ( 30  Q1  Q2 )
L( Q1 ,Q2 , )
 8Q1  Q2    0
Q1
L( Q1 ,Q2 , )
 Q1  10Q2    0
Q2
L( Q1 ,Q2 , )
 Q1  Q2  30  0

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Lecture 2
Solving
8Q1  Q2    0
 Q1  10Q2    0
For Q1, Q2 and λ
 Q1  Q2  30  0
We get:
Q1=16.5
Q2=13.5
λ= 118.5
What does λ mean?
It means that if the constraint were to be relaxed so that more than 30
units could be produced, the cost of producing the 31st is $118.5
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Lecture 2
Example 2:
Imagine that you are running a manufacturing plant. This plant has the
capacity of making 30 units of either widgets or gadgets. Furthermore, the
total cost of the manufacturing operation is:
2
2
C  4G  5W  GW
How many widgets and how many gadgets should you manufacture to
minimize cost?
To minimize cost, we must find the minimum of the cost function above.
We also must make sure that the total units manufactures equals 30.
As such:
G  W  30
therefore
G  30  W
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Lecture 2
Substituting:
G  30  W
Into:
C  4G 2  5W 2  GW
C  4( 30  W )2  5W 2  ( 30  W )W
 10W 2  270W  3600
Taking the derivative and setting it to zero, we get:
dC
 20W  270  0
dW
or
W  13.5
G  30  13.5  16.5
To make sure this is a minimum point:
d dC
(
)  20  0
dW dW
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Lecture 2
Market Demand and the Demand Function
Market Demand Schedule for laptops
Price per unit ($)
Quantity demanded per year (‘000)
3000
800
2750
975
2500
1150
2250
1325
2000
1500
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Lecture 2
Demand Curve
Price
3000
2500
2000
800
1000
1200
1400
1600
Quantity
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Lecture 2
Influences on Demand
Price
Price
3000
2500
2000
3000
2500
Increase in
customer
preference for
laptops
8
0
0
1
0
0
0
Price
1
2
0
0
2000
1
4
0
0
1
6
0
0
8
0
0
Quantity
3000
Increase in
customer per
capita income
1
0
0
0
1
2
0
0
1
4
0
0
1
6
0
0
1
2
0
0
1
4
0
0
1
6
0
0
Quantity
Price
3000
2500
2000
Increase in
advertising for
laptops
8
0
0
1
0
0
0
1
2
0
0
2500
2000
1
4
0
0
1
6
0
0
Quantity
reduction in
cost of
software
8
0
0
1
0
0
0
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Quantity
Lecture 2
Demand Function
Q=f( price of X, Income of consumer, taste of consumer,
advertising expenditure, price of associated goods,….)
Example:
Demand for laptops in 2007 is estimated to
be:
Q= -700P+200I-500S+0.01A
where
P is the average price of laptops in 2007
I is the per capita disposable income in 2007
S is the average price of typical software packages in 2007
A is the average expenditure on advertising in 2007
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Lecture 2
Now let us assume that in 2007:
I=$33,000
S=$400 and
A=$50,000,000
What will be the relationship between price and quantity demanded?
Given that:
Q= -700P+200I-500S+0.01A
We have:
Q= -700P+200(33,000)-500(400)+0.01(50,000,000)
Q= -700P+6,900,000
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Lecture 2
Price Elasticity of Demand
By what percentage would the quantity demanded change as a result of
one unit of change in price?
The percentage change of quantity would be:
Q
Q
The percentage change of price would be:
P
P
Q P
Dividing one by the other:
Q

P
P Q
E ( )
Q P
Rearranging:
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Lecture 2
At the limit:
Therefore:
Q dQ

P dP
E (
P Q
)
Q P
becomes
 (
P dQ
)
Q dP
Example:
Determine the price elasticity of demand for laptops in 2007 when price
is $3000.
We know that:
Q= -700P+6,900,000
dQ
 700
dP
Q=-700(3000)+6900,000=4,800,000
P
3000

 0.000625
Q 4800000
  700  0.000625  0.4375
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Lecture 2
P
P = -aQ+b
  
as
Q 0
b
  1
Demand is price elastic
  1
  1
Demand is price inelastic
 0
b/a
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as
Q
P 0
Lecture 2
Exercise: Show that the price
elasticity of demand on a demand curve
k
given by the equation Q  P is always   1
dQ P
( )
dP Q
1
Qk
p
dQ
1
 k 2
dP
P
P
P P2
P 
Q
k
k

P
Demand
Curve
1 P 2  kP 2
  k 2 

P
k
kP 2
  1
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Q
Lecture 2
Exercise: Given the price elasticity of demand and the price, find marginal
revenue
TR  PQ
dTR dPQ
MR 

dQ
dQ
dQ
dP
dP
P
Q
 P Q
dQ
dQ
dQ
Q dP 

 P 1  ( )(
)
P dQ 

 1
MR  P 1  
 
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Lecture 2
Exercise: Given price elasticity of demand and marginal cost, what is the
maximum price we should charge?
We said that:
 1
MR  P 1  
 
We also know that in order for price to be maximum, MR=MC, so
 1
MR  P 1    MC
 
 1
MC  P 1  
 
or
for
P
P max
MC MC

1  1
1
is the maximum price you should charge

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Lecture 2
Income Elasticity of Demand
By what percentage would the quantity demanded change as a result of
one unit of change in consumer income?
Q
The percentage change of quantity would be:
Q
I
I
The percentage change of income would be:
Q I
Dividing one by the other:
Q
Rearranging:
At the limit:
Therefore:
E (
Q dQ

I
dI
E (
I Q
)
Q I
becomes

I
I Q
)
Q I
I  (
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I dQ
)
Q dI
Lecture 2
Example:
Given that:
Q= -700P+200I-500S+0.01A
Determine the income elasticity of demand for laptops in 2007 when Income is
$33000 S=$400 P=$3000 and A=$50,000,000
dQ
 200
dI
I  (
I dQ
)
Q dI
I  (
I dQ
33000
)

 200  1.375
Q dI 4800000
Therefore one percent increase in income leads to 1.375 percent
increase in demand for laptops.
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Lecture 2
Cross Elasticity of Demand
By what percentage would the quantity demanded change as a result of
one unit of change in the price of an associated product?
Example:
 XY
PY dQ X
(
)
Q X dPY
Determine the cross elasticity of demand for laptops in 2007 when price of
software is $400
 XY  (
PY dQ X
400
)

 500  0.042
Q X dPY 4800000
Therefore one percent increase in price of software leads to 0.042
percent decrease in demand for laptops.
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