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1
2. Economic Applications of Single-Variable Calculus
• Derivative Origins• Single Variable Derivatives• Economic uses of Derivatives
2
2. Economic Applications of Single-Variable Calculus
2.1 Derivatives of Single-Variable Functions
2.2 Applications using Derivatives
3
2. Economic Applications of Single-Variable Calculus
In economics, derivatives are used in various ways:
Marginal amounts (slope) Maximization Minimization Graphing Elasticity and Growth
4
2.1 Derivatives of Single-Variable Functions
Slope:
-consider the following graph of demand
-what is the impact of a $55 drop in price?
5
2.1 – Slope
Slope Approximation:
1 2 3 4 5 6 7 80
20
40
60
80
100
120
P=100-40ln(Q)
x
Quantity Demanded
Pri
ce
Slope = a decrease of $45 increases demand by 3, or $15 price decrease per quantity increase (slope).
6
What is the impact of changing a price of $35?
Slope = rise/run
=Δq/ Δp = (q1-q0)/(p1-p0)
1 2 3 4 5 6 7 8 9 100.00
20.00
40.00
60.00
80.00
100.00
120.00
Demand for Board Games
Series1
Quantity
Pri
ce
Slope AB =(q1-q0)/(p1-p0)
=(35-100)/(3-1)
=-65/2
=-32.5Slope BC =(q1-q0)/(p1-p0)=(20-35)/(5-3)=-15/2=-7.5But what is the impact of a SMALL price change at B?
7
2.1 Derivatives of Single-Variable Functions
--in order to find the slope AT B (the impact of a small price change), one must find an INSTANTANEOUS SLOPE
-slope of a tangent line at B
-derivative at B
8
2.1 – Tangents
The green tangent line represents the instantaneous slope
Slopes
0.00
20.00
40.00
60.00
80.00
100.00
120.00
1 2 3 4 5 6 7 8 9 10
Price
Qu
an
tity
Series1
9
2.1 Instantaneous Slope
To calculate an instantaneous slope/find a derivative (using calculus), you need:
1) A function
2) A continuous function
3) A smooth continuous function
10
2.1.1 – A FunctionDefinitions:
-A function is any rule that assigns a maximum and minimum of one value to a range of another value
-ie y=f(x) assigns one value (y) to each x
-note that the same y can apply to many x’s, but each x has only one y
-ie: y=x1/2 is not a function
x = argument of the function (domain of function)
f(x) or y = range of function
11
Function:
Each X
Has 1 Y
y=0+2sin(2pi*x/14)+2cos(2pi*x/14)
-4
-3
-2
-1
0
1
2
3
4
1 3 5 7 9 11 13 15 17 19
x
y x
12
Not a Function:
Here each x Corresponds to 2 y values
Often called the straight line test
13
2.1.1 – Continuous-if a function f(x) draws close to one finite number L for all values of x as x draws closer to but does not equal a, we say:
lim f(x) = Lx-> a
A function is continuous iff (if and only iff)
i) f(x) exists at x=aii) Lim f(x) exists x->aiii) Lim f(x) = f(a) x->a
14
2.1.1 – Limits and ContinuityIn other words:
i) The point must exist
ii) Points before and after must exist
iii)These points must all be joined
Or simply:
The graph can be drawn without lifting one’s pencil.
15
2.1.2 Smooth-in order for a derivative to exist, a function must be continuous and “smooth” (have only one tangent)
Slopes
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
1 2 3 4 5 6 7 8 9 10
Price
Qu
an
tity
Series1
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2.1.2 Derivatives
-if a derivative exists, it can be expressed in many different forms:
a)dy/dx
b)df(x)/dx
c) f ’(x)
d)Fx(x)
e)y’
17
2.1.2 Derivatives and Limits
-a derivative (instantaneous slope) is derived using limits:
x=0+2sin(2pi*t/14)+2cos(2pi*t/14)
-4
-3
-2
-1
0
1
2
3
4
1 3 5 7 9 11 13 15 17 19
t
x x
This method is known as differentiation by first principles, and determines the slope between A and B as AB collapses to a point (A)
))()(
(lim)('0 h
xfhxfxf
h
18
2.1.2 Rules of Derivatives
-although first principles always work, the following rules are more economical:
1) Constant Rule
If f(x)=k (k is a constant),f ‘(x) = 0
2) General Rule
If f(x) = ax+b (a and b are constants)f ‘ (x) = a
19
2.1.2 Examples of Derivatives
1) Constant Rule
If f(x)=27 f ‘(x) = 0
2) General Rule
If f(x) = 3x+12 f ‘(x) = 3
20
2.1.2 Rules of Derivatives
3) Power Rule
If f(x) = kxn,f ‘(x) = nkxn-1
4) Addition Rule
If f(x) = g(x) + h(x),f ‘(x) = g’(x) + h’(x)
21
2.1.2 Examples of Derivatives
3) Power Rule
If f(x) = -9x7,f ‘(x) = 7(-9)x7-1
=-63x6
4) Addition Rule
If f(x) = 32x -9x2
f ‘(x) = 32-18x
22
2.1.2 Rules of Derivatives
5) Product Rule
If f(x) =g(x)h(x),f ‘(x) = g’(x)h(x) + h’(x)g(x)-order doesn’t matter
6) Quotient Rule
If f(x) =g(x)/h(x),f ‘(x) = {g’(x)h(x)-h’(x)g(x)}/{h(x)2}-order matters-derived from product rule (implicit derivative)
23
2.1.2 Rules of Derivatives
5) Product Rule
If f(x) =(12x+6)x3
f ‘(x) = 12x3 + (12x+6)3x2
= 48x3 + 18x2
6) Quotient Rule
If f(x) =(12x+1)/x2
f ‘(x) = {12x2 – (12x+1)2x}/x4
= [-12x2-2x]/x4
= [-12x-2]/x3
24
2.1.2 Rules of Derivatives
7) Power Function Rule
If f(x) = [g(x)]n,f ‘(x) = n[g(x)]n-1g’(x)-work from the outside in-special case of the chain rule
8) Chain Rule
If f(x) = f(g(x)), let y=f(u) and u=g(x), thendy/dx = dy/du X du/dx
25
2.1.2 Rules of Derivatives
7) Power Function Rule
If f(x) = [3x+12]4,f ‘(x) = 4[3x+12]33
= 12[3x+12]3
8) Chain Rule
If f(x) = (6x2+2x)3 , let y=u3 and u=6x2+2x, dy/dx = dy/du X du/dx
= 3u2(12x+2) = 3(6x2+2x)2(12x+2)
26
2.1.2 More Exciting Derivatives
1) Inverses
If f(x) = 1/x= x-1,f ‘(x) = -x-2=-1/x2
1b) Inverses and the Chain Rule
If f(x) = 1/g(x)= g(x)-1,f ‘(x) = -g(x)-2g’(x)=-1/g(x)2g’(x)
27
2.1.2 – More Exciting Derivatives
2) Natural Logs
If y=ln(x),
y’ = 1/x
-chain rule may apply
If y=ln(x2)
y’ = (1/x2)2x = 2/x
28
2.1.2 – More Exciting Derivatives
3) Trig. Functions
If y = sin (x),
y’ = cos(x)
If y = cos(x)
y’ = -sin(x)
-We see this relationship graphically:
29
2.1.2 – More Derivatives
Reminder: derivatives reflect slope: Sine(blue) and Cosine(red)
-1.5
-1
-0.5
0
0.5
1
1.5
1 3 5 7 9 11 13 15 17 19 21 23 25 27
x
sin
(x),
co
s(x
)
30
2.1.2 – More Derivatives
3b) Trig. Functions – Chain Rule
If y = sin2 (3x+2),
y’= 2sin(3x+2)cos(3x+2)3
Exercises:
y=ln(2sin(x) -2cos2(x-1/x))
y=sin3(3x+2)ln(4x-7/x3)5
y=ln([3x+4]sin(x)) / cos(12xln(x))
31
2.1.2 – More Derivatives
4) Exponents
If y = bx
y’ = bxln(b)
Therefore
If y = ex
y’ = ex
32
2.1.2 – More Derivatives
4b) Exponents and chain rule
If y = bkx
y’ = bkxln(b)k
Or more generally:
If y = bg(x)
y’ = bg(x)ln(b) X dg(x)/dx
33
2.1.2 – More Derivatives
4b) Exponents and chain rule
If y = 52x
y’ = 52xln(5)2
Or more complicated:
If y = 5sin(x)
y’ = 5sin(x)ln(5) * cos(x)
34
2.1.2.1 – Higher Order Derivatives
-First order derivates (y’), show us the slope of a graph
-Second order derivatives measure the instantaneous change in y’, or the slope of the slope
-or the change in the slope:
-(Higher-order derivates are also possible)
35
Here the slopeincreases as tincreases, transitioningfrom a negativeslope to a positiveslope.
A second derivative would be positive, and confirm a minimum point on the graph.
2.1.2.1 Second Derivativesx=15-10t+t*t
-15
-10
-5
0
5
10
1 2 3 4 5 6 7 8
t
x x
36
x=15+10t-t*t
0
10
20
30
40
50
1 2 3 4 5 6 7 8
t
x x
Here, the slope moves from positive to negative, decreasingover time.
A second derivative would be negative and indicate a maximum point on the graph.
2.1.2.1 Second Derivatives
37
2.1.2.1 – Second Order DerivativesTo take a second order derivative:
1) Apply derivative rules to a function
2) Simplify if possible
3) Apply derivative rules to the answer to (1)
Second order differentiation can be shown a variety of ways:
a)d2y/dx2 b) d2f(x)/dx2
c)f ’’(x) d) fxx(x)
e) y’’
38
2.1.2.1 – Second Derivative Examples
y=12x3+2x+11y’=36x2+2y’’=72x
y=sin(x2)y’=cos(x2)2xy’’=-sin(x2)2x(2x)+cos(x2)2y’’’=-cos(x2)2x(4x2)-sin(x2)8x-sin(x2)2x(2) =-cos(x2)8x3-sin(x2)12x
39
2.1.2.2 – Implicit Differentiation
So far we’ve examined cases where our function is expressed:
y=f(x) ie: y=7x+9x2-14
Yet often equations are expressed:
14=7x+9x2-y
Which requires implicit differentiation.
-In this case, y can be isolated. Often, this is not the case
40
2.1.2.2 – Implicit Differentiation Rules
1) Take the derivative of EACH term on both sides.
2) Differentiate y as you would x, except that every time you differentiate y, you obtain dy/dx (or y’)
Ie: 14=7x+9x2-yd(14)/dx=d(7x)/dx+d(9x2)/dx-dy/dx0 = 7 + 18x – y’y’=7+18x
41
2.1.2.2 – Implicit Differentiation Examples
Sometimes isolating y’ requires algebra:
xy=15+x
y+xy’=0+1
xy’=1-y
y’=(1-y)/x (this can be simplified to remove y)
= [1-(15+x)/x]/x
= (x-15-x)/x2
=(-15)/x2
42
2.1.2.2 – Implicit Differentiation Examples
x2-2xy+y2=1
d(x2)/dx+d(2xy)/dx+d(y2)/dx=d1/dx
2x-2y-2xy’+2yy’=0
y’(2y-2x)=2y-2x
y’=(2y-2x) / (2y-2x)
y’=1
43
2.1.2.2 – Implicit Differentiation Examples
Using the implicit form has advantages:3x+7y8=183+56y7y’=056y7y’=-3y’=-3/56y7
vrs.y=[(18-3x)/7)1/8
y’=1/8 * [(18-3x)/7)-7/8 * 1/7 * (-3)Which simplifies to the above.
44
2.2.1 Derivative Applications - Graphs
Derivatives can be used to sketch functions:First Derivative:-First derivative indicates slope-if y’>0, function slopes upwards-if y’<0, function slopes downwards-if y’=0, function is horizontal-slope may change over time
-doesn’t give shape of graph
45
2.2.1 Positive Slope Graphs
0
20
40
60
80
100
120
1 2 3 4 5 6 7 8 9 10
Linear, Quadratic, and Lin-Log Graphs
46
2.2.1 Derivative Applications - Graphs
Next, shape/concavity must be determined
Second Derivative:
-Second derivative indicates concavity
-if y’’>0, slope is increasing (convex)
-if y’’<0, slope is decreasing (concave, like a hill or a cave)
-if y’’=0, slope is constant (or an inflection point occurs, see later)
47
2.2.1 Sample Graphs
x’’= 2, slope is increasing; graph is convex
x=15-10t+t*t
-15
-10
-5
0
5
10
1 2 3 4 5 6 7 8
t
x x
48
2.2.1 – Sample Graphs
x’’=-2, slope is decreasing; graph is concave
x=15+10t-t*t
0
10
20
30
40
50
1 2 3 4 5 6 7 8
t
x x
49
2.2.1 Derivative Applications - Graphs
Maxima/minima can aid in drawing graphsMaximum Point:If 1) f(a)’=0, and
2) f(a)’’<0, -graph has a maximum point (peak) at x=aMinimum Point:If 1) f(a)’=0, and
2) f(a)’’>0, -graph has a minimum point (valley) at x=a
50
2.2.1 Sample Graphs
x’’= 2, slope is increasing; graph is convex
x=15-10t+t*t
-15
-10
-5
0
5
10
1 2 3 4 5 6 7 8
t
x x
x’=-10+2t=0t=5
51
2.2.1 – Sample Graphs
x’’=-2, slope is decreasing; graph is concave
x=15+10t-t*t
0
10
20
30
40
50
1 2 3 4 5 6 7 8
t
x x
x’=10-2t=0t=5
52
2.2.1 Derivative Applications - Graphs
Inflection Points:
If 1) f(a)’’=0, and
2) the graph is not a straight line
-then an inflection point occurs
-(where the graph switches between convex and concave)
53
2.2.1 Derivatives and Graphing
Cyclical case:
x=0+2sin(2pi*t/14)+2cos(2pi*t/14)
-4
-3
-2
-1
0
1
2
3
4
1 3 5 7 9 11 13 15 17 19
t
x x
54
2.2.1 Derivative Applications - Graphs
7 Graphing Steps:
i) Evaluate f(x) at extreme points (x=0, ∞, - ∞, or a variety of values)
ii) Determine where f(x)=0
iii) Calculate slope: f ’(x) - and determine where it is positive and negative
iv) Identify possible maximum and minimum co-ordinates where f ‘(x)=0. (Don’t just find the x values)
55
2.2.1 Derivative Applications - Graphs
7 Graphing Steps:
v) Calculate the second derivative – f ‘’(x) and use it to determine max/min in iv
vi) Using the second derivative, determine the curvature (concave or convex) at other points
vii) Check for inflection points where f ‘’(x)=0
56
2.2.1 Graphing Example 1y=(x-5)2-3
i) f(0)=22, f(∞)= ∞, f(-∞)=∞
ii) y=0 when
(x-5)2=3
(x-5) = ± 31/2
x = ± 31/2+5
x = 6.7, 3.3 (x-intercepts)
iii) y’=2(x-5)
y’>0 when x>5
y’<0 when x<5
57
2.2.1 Graphing Example 1
y=(x-5)2-3
iv) y’=0 when x=5
f(5)=(5-5)2-3=-3
(5,-3) is a potential max/min
v) y’’=2, (5,-3) is a minimum
vi) Function is always positive, it is always convex
vii) y’’ never equals zero
58
2.2.1 Graphing Example 1
y=(x-5)(x-5)-3
-5
0
5
10
15
20
25
1 2 3 4 5 6 7 8 9 10
x
y
(0,22)
(3.3,0) (6.7,0)
(5,-3)
59
2.2.1 Graphing Example 2y=(x+1)(x-3)=x2-2x-3
i) f(0)=-3, f(∞)= ∞, f(-∞)=∞
ii) y=0 when
(x+1)(x-3) =0
x = 3,-1 (x-intercepts)
iii) y’=2x-2
y’>0 when x>1
y’<0 when x<1
60
2.2.1 Graphing Example 2
y=(x+1)(x-3)=x2-2x-3
iv) y’=0 when x=1
f(1)=12-2(1)-3=-4
(1,-4) is a potential max/min
v) y’’=2, x=1 is a minimum
vi) Function is always positive, it is always convex
vii) y’’ never equals zero
61
y=(x+1)(x-3)
-5, 32
-4, 21
-3, 12
-2, 5-1, 0
0, -31, -42, -33, 0
4, 5
5, 12
6, 21
7, 32
8, 45
9, 60
-10
0
10
20
30
40
50
60
70
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
2.2.1 Graphing Example 2
62
2.2.2 Optimization
Some claim economists have 3 jobs:
1) Analyze what has happened (past)
2) Describe the current economy (present)
3) Advise on future decisions (future)
For #3, an economist must first calculate the best possible result.
63
2.2.2 Optimization
Optimization falls into two categories:
1) Maximization (ie: production, profits, utility, happiness, grades, health, employment, etc.)
2) Minimization (ie: costs, pollution, disutility, unemployment, sickness, homework, etc.)
64
2.2.2 Optimizing in 3 Steps
There are three steps for optimization:1) FIRST ORDER CONDITION (FOC)
Find where f’(x)=0. These are potential maxima/minima.
2) SECOND ORDER CONDITION (SOC) Evaluate f’’(x) at your potential maxima/minima. This determines if (1)’s solutions are maxima/minima/inflection points
3) Co-Ordinates Obtain the co-ordinates of your maxima/minima
65
2.2.2 Optimizing Example 1
Cooking is tricky – too long spent cooking, and it burns, too little time spent cooking – and some of it is raw and inedible.
Production of oatmeal is expressed as:
Let x=15+10t-t2
x = bowls of oatmeal
t = 5 minute intervals of time
Maximize Oatmeal Production
66
2.2.2 Optimous Oatmeal
Let x=15+10t-t2
FOC:
x’ = 10-2t = 0
10 = 2t
5 = t
SOC:
x’’ = -2
x’’ < 0, concave, maximum
67
2.2.2 Optimous Oatmeal
Let x=15+10t-t2
Co-ordinates:
x(5) = 15+10(5)-52
x(5) = 15+50-25
= 40
Gourmet oatmeal production is maximized at 40 bowls when 25 minutes (5X5) are spent cooking (All else held equal).
68
2.2.2 – Oatmeal for everyone
Production is maximized at (5,40).
x=15+10t-t*t
0
10
20
30
40
50
1 2 3 4 5 6 7 8
t
x x
69
2.2.2 Marriage and Motorcycles
Steve wants to buy a new motorbike. Being a married man however, he knows that his utility is directly tied to his wife’s opinion of the idea. Furthermore, he knows it’s best to bring it up to Denise (his wife) when she’s at her weakest.
Denise’s daily opposition to a motorcycle is expressed as x=-cos(tπ/12)
Where t = hour of the day (0-24)
When should Steve ask Denise?
70
2.2.2 M & M
Let x=-cos(tπ/12) FOC:
x’ = sin(tπ/12)π/12 = 0
0 = sin(tπ/12)
This occurs when
t π/12 = 0, π, 2π
t = 0, 12, 24
2 possible mimima (0=24 on the clock)
71
2.2.2 Early Bird Gets the Motorcycle
Let x=-cos(tπ/12)
SOC:
x’’ = cos(tπ/12)(π/12)2
x’’ > 0 when 0≤t<6, 18<t≤24
-convex, min (0, 24 are acceptable)
x’’ < 0 when 6<t<18
-concave, max (12 is out)
0 or 24
6
12
18 +-
72
2.2.2 Early Bird Gets the Motorcycle
Let x=-cos(tπ/12) Results:
x(0)=-1
x(24)=-1
Denise has her least resistance (of -1) at midnight. Steve’s best move is to bring up the motorcycle when Denise is tired or asleep.
73
2.2.2 Early Bird Gets the Motorcycle
Let x=-cos(tπ/12)
Denise's Resistance to Motorcycles
-1.5
-1
-0.5
0
0.5
1
1.5
Time (Hours of the day)
Re
sis
tan
ce
74
2.2.2 Necessary and Sufficient
The FOC provides a NECESSARY condition for a maximum or minimum.
The FOC is not a SUFFICIENT condition for a maximum or minimum.
The FOC and SOC together are NECESSARY AND SUFFICIENT conditions for a max. or min.
75
2.2.2 Marginal Concepts
Marginal Profit = Change in profit from the sale/production of one extra unit.
MP=dπ/dq
If Marginal Profit >0, quantity should increase; as the next unit will increase profit.
If Marginal Profit <0, quantity should decrease, as the last unit decreased profit
76
2.2.2 Marginal Concepts
Quantity is therefore optimized when Marginal Profit=0;
Ie: when dπ/dq=0.
SOC still confirms that this is a maximum (ie: that the previous unit increased profit and the next unit will decrease profit.)
77
2.2.2 Marginal Concepts
Alternately, remember that
Profit = Total Revenue – Total Costs
Or
π = TR-TC
therefore
Mπ=MR-MC (dπ/dq=dTR/dq-dTC/dq)
So Mπ = 0 is equivalent to saying that
MR-MC =0
MR=MC
78
2.2.2 The Outfit Example
Your significant other shows you a new, horrible outfit they just bought and asks how they look. You think of 6 possible lies:
1) You look amazing
2) You have such great taste
3) That colour really brings out your eyes
4) Neon is in this year
5) I should get a matching outfit
6) You should wear that to my office party
79
2.2.2 The Outfit Example
The benefit lying is
TR=2L3+60L
The cost of lying is
TC=21L2+100
Where L = number of lies
How many lies should you tell?
80
2.2.2 The Outfit Example
Profit=TR-TC
Profit=2L3+60L-[21L2+100]
Profit=2L3-21L2+60L-100
FOC:
Mπ=6L2-42L+60
Mπ=6(L2-7L+10)
81
2.2.2 The Outfit Example
Mπ=6(L2-7L+10)
Solving using the quadratic formula:
L=-b±(b2-4ac)1/2 / 2a
L=-7±(49-40)1/2/(2)
L=(-7±3)/2 = 2, 5
82
2.2.2 The Outfit ExampleProfit=2L3-21L2+60L-100
Mπ=6L2-42L+60
SOC: π’’=12L-42
π’’(2)=12(2)-42=-18, concave MAX
π’’(5)=12(5)-42=18, convex MIN
You should tell 2 lies
83
2.2.2 The Outfit ExampleProfit=2L3-21L2+60L-100
Profit(2)=2(2)3-21(2)2+60(2)-100
Profit(2)=16-84+120-100
Profit(2)=-48
You are minimizing the damage at -48 by telling two lies.
84
2.2.2 Constrained OptimizationThus far we have considered UNCONSTRAINED optimization. (No budget constraint, time constraint, savings constraint, etc.)
We can also deal with CONSTRAINED optimization.
Ie: Maximize utility with respect to a set income.
Maximize income with respect to a 24 hour day.
Maximize lie effectiveness while keeping a straight face.
To use this, we use a Lagrangean. (chapter 4)
85
2.2.3 Elasticities
We have already seen how the derivative, or the slope, can change as x and y change
-even if a slope is constant, changes can have different impacts at different points
-For example, given a linear demand for Xbox 720’s, a $100 price increases affects profits differently at different starting prices:
86
2.2.3 Xbox 720 Example
Price increase from $0 to $100
Xbox Demand
0
1000000
2000000
3000000
4000000
5000000
6000000
0 100 200 300 400 500 600 700 800 900
Xbox Price
Xb
ox
De
ma
nd
New
In
com
e
87
2.2.3 Xbox 720- Example
Price increase from $500 to $600
Xbox Demand
0
1000000
2000000
3000000
4000000
5000000
6000000
0 100 200 300 400 500 600 700 800 900
Xbox Price
Xb
ox
De
ma
nd
Old Income
New Income
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2.2.3 Elasticities$0 to $100
Old Revenue: $0
New Revenue: 4.5 million sold X $100 each
$450 million (INCREASE)
$500 to $600
Old Revenue: 2.5 million sold X $500 each
$1.25 Billion
New Revenue: 1.5 million sold X 600 each
$0.9 Billion (DECREASE)
89
2.2.3 Elasticities-to avoid this problem, economists often utilize
ELASTICITIES
-elasticities deal with PERCENTAGES and are therefore more useful across a variety of points on a curve
ELASTICITY = a PROPORTIONAL change in y from a PROPORTIONAL change in x
Example: elasticity of demand:
E = Δq/q / Δp/p
= (Δq/Δp) (p/q)
= (dq/dp) (p/q)
90
2.2.3 Elasticity Example 1Let Q=12P-7
Find elasticities at x= 5, and 10
1) dq/dp = 12
2) q(5) = 12(5) - 7 = 53
3) q(10) = 12(10) -7 = 113
Next we apply the formula:
91
2.2.3 Elasticity Example 1Let Q=12P-7
Find elasticities at (p,q)= (5,53) and (10,113)
E = dq/dp * p/q
1) E (5)= 12 * 5/53 = 1.13
2) E (10)= 12 * 10/113 = 1.06
92
2.2.3 Elasticity Interpretation
What does an elasticity of X mean?
=> for a 1% increase in x (or the independent variable), there will be a X% increase in y (or the dependent variable)
In our example, a 1% increase in P caused a:1)1.13 % increase in Q
2)1.06 % increase in Q
Question: Is an increase in Q good or bad?
93
2.2.3 Inelastic vrs. Elastic: The Boxer’s or Briefs debate
An elasticity of less than 1 (in absolute terms) is inelastic.
That is, y responds less than x in percentage terms.
An elasticity of greater than 1 (in absolute terms) is elastic.
That is, y responds more than x in percentage terms.
-This has policy implications…
94
2.2.3 Elastic XboxesLet q=5,000,000-5,000p Or
q=5,000-5p
Where q is Xbox’s demanded in 1000’s
p is price of an Xbox
Find elasticities at p=0, 100, and 500
1) dq/dp = -5
2) Q(0) =5,000
3) Q(100)=4,500
4) Q(500)=2,500
95
2.2.3 Elastic XboxesFind elasticities at p,q =(0,5000), (100,4500)
and (500,2500)
E = dq/dp * p/q
1) E = -5* 0/5000 = 0
2) E = -5 * 100/4500 = -0.11 (inelastic)
3) E = -5 * 500/2500 = -1 (unit elastic)
96
2.2.3 Elastic XboxesFrom these values, we know that demand for
Xboxes 720’s is INELASTIC below $500 and ELASTIC above $500
How does this impact revenue?
Total Revenue = p*q(p)
dTR/dp = q(p)+p*dq/dp (chain rule)
= q( 1+p/q*dq/dp)
= q (1+ E)
97
2.2.3 Making Microsoft MoneyIf E = -1, dTR/dp = 0; a small change in price
won’t affect revenue
If |E| < 1 (inelastic), dTR/dp>0, small increases in prices increase revenue
If |E| > 1 (elastic), dTR/dp<0, small increases in prices decrease revenue
Therefore, price increases are revenue enhancing up to a price of $500.
98
2.2.3 Assuming costs are constant:
If demand is inelastic
Raise Price
If demand is elastic
Decrease Price
If demand is unit elastic
Price is perfect (usually)
99
2.2.3 More Elasticity ExercisesLet q = 100-2p
1) Find Elasticities at p=5, 20 and 40
2) Formulate Economic Advice at these points
Let q = 200+2p-4p2
3) Find Elasticities at p=0, 5 and 10
4) Formulate Economic Advice at these points
100
2.2.3 Elastic LogsANOTHER reason to use logs in economic
formulae is to more easily calculate elasticities:
E = dy/dx * x/y= (1/y) dy/dx (x)= (dlny/dy) dy/dx (dx/dlnx)= (dlny/dlnx) dx/dx (dy/dy)= (dlny/dlnx)
101
2.2.3 Examples are a log’s best friend
Let ln(q) = -1ln(p)
E = dln(q)/dln(p)= -1
Hence demand is unit elastic and change in price would not affect total revenue.
102
2.2.3 Log Elasticity Exercises
Let ln(q) = 100+ln(30/p)
1) Find Elasticities at p=5, 10 and 20
2) Formulate Economic Advice at these points
Let ln(q) = 1/2 ln(p2)
3) Find Elasticities at p=0.5, 1, and 2
103
3.1.1 – Growth Formulas
Calculating growth between 2 points:
1) [{Xt-Xt-1}/Xt-1] X 100
2) [ln(Xt)-ln(Xt-1)] X 100
Calculating instantaneous growth using derivatives:
3) [{dX/dt}/X] X 100
4) [dln(X)/dt] X 100
104
3.1.1 – Growth Derivation
gdt
Xd
tgXX
gtXX
XgX
XgX
XXXg
tt
tt
tt
tt
t
tt
)ln(
)ln()ln(
)1ln()ln()ln(
)1(
)1(
)(
1
1
0
1
1
1
Proof of second formula from the chain rule,
X
dtdX
dt
dX
xdt
dX
dX
Xd
dt
Xd
1lnln
105
3.1.1 – Math Growth Example 1
a) Find growth of X when ln(X) = 10 +2t
Using 4:
g = dln(X)/dt
= 2 (or 200%)
b) Find growth of X when t = 4
g(t=4) = 200% (since growth is always 200% for this formula)
106
3.1.1 – Math Growth Example 2
a) Find growth of X when ln(X) = 10 - 2ln(t)
Using 4:
g = dln(X)/dt
= -2/t (or -200%/t)
b) Find growth of X when t=20
g (t=20) = (-200/20)%
g (t=20) = 10%
107
3.1.1 – Math Growth Example 2
Growth - % change in y when x increases by 1
Y
dXdY
dX
Ydg
ln
Elasticity - % change in y when x increases by 1 %
Y
X
dX
dY
Xd
Yd
ln
ln