Transcript Econ 299 Chapter 02

2. Economic
Applications of SingleVariable Calculus
• Derivative Origins
• Single Variable Derivatives
• Economic uses of Derivatives
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2. Economic Applications of SingleVariable Calculus
2.1 Derivatives of SingleVariable Functions
2.2 Applications using
Derivatives
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2. Economic Applications of Single-Variable
Calculus
In economics, derivatives are used in
various ways:
 Marginal amounts (slope)
 Maximization
 Minimization
 Graphing
 Elasticity and Growth
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2.1 Derivatives of Single-Variable Functions
Slope:
-The impact on the “y” variable when the
“x” variable increases by 1
-consider the following graph
-what is the impact of changing how
much time you spend on assignment
1?
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What is the impact of changing a price?
Slope AB
=(y1-y0)/(x1-x0)
Slope = rise/run
=Δy/ Δx = (y1-y0)/(x1-x0)
Assignment 1 Mistakes
=-32.5
Slope BC
=(y1-y0)/(x1-x0)
=(20-35)/(5-3)
=-7.5
120.00
100.00
Mistakes
=(100-35)/(1-3)
80.00
60.00
40.00
20.00
0.00
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2
3
4
5
6
7
8
9 10
Hours Spent on Assignment 1
But what is the impact
of a SMALL time
change at 3 hours?
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2.1 Derivatives of Single-Variable Functions
--in order to find the slope AT B (the
impact of a small time change), one
must find an INSTANTANEOUS
SLOPE
-slope of a tangent line at B
-derivative at B
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2.1 – Tangents
Slopes
120.00
Quantity
100.00
80.00
60.00
Series1
40.00
20.00
0.00
1
2
3
4
5
6
7
8
9 10
Price
The green tangent line represents the
instantaneous slope
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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
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2.1.1 – A Function
Definitions:
-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
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Function:
y=0+2sin(2pi*x/14)+2cos(2pi*x/14)
y
Each X
4
3
2
1
0
-1
-2
-3
-4
Has 1 Y
x
1
3
5
7
9
11
13
15
17
19
x
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Not a Function:
Here each x
Corresponds
to 2 y values
Often called the straight line
test
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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) = L
x-> a
A function is continuous iff (if and only iff)
i) f(x) exists at x=a
ii) Lim f(x) exists
x->a
iii) Lim f(x) = f(a)
x->a
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2.1.1 – Limits and Continuity
In 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.
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2.1.2 Smooth
-in order for a derivative to exist, a function must be
continuous and “smooth” (have only one tangent)
Slopes
40.00
35.00
Quantity
30.00
25.00
20.00
Series1
15.00
10.00
5.00
0.00
1
2
3
4
5
6
Price
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8
9 10
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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’
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2.1.2 Derivatives and Limits
-a derivative (instantaneous slope) is derived
using limits:
f ( x  h)  f ( x )
f ' ( x)  lim (
h 0
x=0+2sin(2pi*t/14)+2cos(2pi*t/14)
4
3
2
x
1
0
-1
x
1
3
5
7
9
11
-2
-3
-4
t
13
15
17
19
h
)
This method is known
as differentiation by
first principles, and
determines the slope
between A and B as
AB collapses to a
point (A)
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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
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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
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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)
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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
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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)
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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
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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), then
dy/dx = dy/du X du/dx
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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)
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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)
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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
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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:
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2.1.2 – More Derivatives
Reminder: derivatives reflect slope:
Sine(blue) and Cosine(red)
1.5
sin(x),cos(x)
1
0.5
0
-0.5
1
3
5
7
9 11 13 15 17 19 21 23 25 27
-1
-1.5
x
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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))
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2.1.2 – More Derivatives
4) Exponents
If y = bx
y’ = bxln(b)
Therefore
If y = ex
y’ = ex
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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
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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)
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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)
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2.1.2.1 Second Derivatives
x=15-10t+t*t
10
5
0
x
Here the slope
increases as t
increases,
transitioning
from a negative
slope to a positive
slope.
-5
1
2
3
4
5
6
7
x
8
-10
-15
t
A second derivative would be positive, and confirm
a minimum point on the graph.
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2.1.2.1 Second Derivatives
x=15+10t-t*t
50
40
30
x
x
Here, the slope
moves from
positive to
negative,
decreasing
over time.
20
10
0
1
2
3
4
5
6
7
8
t
A second derivative would be negative and indicate
a maximum point on the graph.
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2.1.2.1 – Second Order Derivatives
To 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
c)f ’’(x)
e) y’’
b) d2f(x)/dx2
d) fxx(x)
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2.1.2.1 – Second Derivative Examples
y=12x3+2x+11
y’=36x2+2
y’’=72x
y=sin(x2)
y’=cos(x2)2x
y’’=-sin(x2)2x(2x)+cos(x2)2
y’’’=-cos(x2)2x(4x2)-sin(x2)8x-sin(x2)2x(2)
=-cos(x2)8x3-sin(x2)12x
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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
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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-y
d(14)/dx=d(7x)/dx+d(9x2)/dx-dy/dx
0 = 7 + 18x – y’
y’=7+18x
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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
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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
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2.1.2.2 – Implicit Differentiation Examples
Using the implicit form has advantages:
3x+7y8=18
3+56y7y’=0
56y7y’=-3
y’=-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.
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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
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2.2.1 Positive Slope Graphs
Linear, Quadratic, and Lin-Log Graphs
120
100
80
60
40
20
0
1
2
3
4
5
6
7
8
9
10
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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)
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2.2.1 Sample Graphs
x’’= 2, slope is increasing; graph is convex
x=15-10t+t*t
10
5
x
0
-5
1
2
3
4
5
6
7
8
x
-10
-15
t
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2.2.1 – Sample Graphs
x’’=-2, slope is decreasing; graph is concave
x=15+10t-t*t
50
40
30
x
x
20
10
0
1
2
3
4
5
t
6
7
8
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2.2.1 Derivative Applications - Graphs
Maxima/minima can aid in drawing graphs
Maximum Point:
If
1) f(a)’=0, and
2) f(a)’’<0,
-graph has a maximum point (peak) at x=a
Minimum Point:
If
1) f(a)’=0, and
2) f(a)’’>0,
-graph has a minimum point (valley) at x=a
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2.2.1 Sample Graphs
x’’= 2, slope is increasing; graph is convex
x=15-10t+t*t
x’=-10+2t=0
t=5
10
5
x
0
-5
1
2
3
4
5
6
7
8
x
-10
-15
t
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2.2.1 – Sample Graphs
x’’=-2, slope is decreasing; graph is concave
x=15+10t-t*t
x’=10-2t=0
t=5
50
40
30
x
x
20
10
0
1
2
3
4
5
t
6
7
8
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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)
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2.2.1 Derivatives and Graphing
Cyclical case:
x=0+2sin(2pi*t/14)+2cos(2pi*t/14)
4
3
2
x
1
0
-1
x
1
3
5
7
9
11
13
15
17
19
-2
-3
-4
t
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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)
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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
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2.2.1 Graphing Example 1
y=(x-5)2-3
f(0)=22, f(∞)= ∞, f(-∞)=∞
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
i)
ii)
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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
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2.2.1 Graphing Example 1
y=(x-5)(x-5)-3
25
20
(0,22)
y
15
10
5
(3.3,0)
(6.7,0)
0
-5
1
2
3
4
(5,-3)
5
6
7
8
9
10
x
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2.2.1 Graphing Example 2
y=(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
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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
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2.2.1 Graphing Example 2
y=(x+1)(x-3)
70
60
9, 60
50
8, 45
40
30
20
10
0
-10
-5, 32
-4, 21
7, 32
6, 21
-3, 12
5, 12
-2, 5
4, 5
-1, 0
3, 0
0, -31, -42, -3
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
8
9
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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.
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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.)
62
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
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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
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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
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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).
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2.2.2 – Oatmeal for everyone
Production is maximized at (5,40).
x=15+10t-t*t
50
40
30
x
x
20
10
0
1
2
3
4
5
t
6
7
8
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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?
68
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)
69
2.2.2 Early Bird Gets the Motorcycle
0 or 24
Let x=-cos(tπ/12)
18
+
-
6
SOC:
x’’ = cos(tπ/12)(π/12)2
12
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)
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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.
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2.2.2 Early Bird Gets the Motorcycle
Let x=-cos(tπ/12)
Denise's Resistance to Motorcycles
1.5
0.5
24
22
20
18
16
14
12
10
8
6
4
-0.5
2
0
0
Resistance
1
-1
-1.5
Time (Hours of the day)
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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.
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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
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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.)
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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
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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
77
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?
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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)
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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
80
2.2.2 The Outfit Example
Profit=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
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2.2.2 The Outfit Example
Profit=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.
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2.2.2 Constrained Optimization
Thus 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. (chapter834)
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:
84
2.2.3 Xbox 720 Example
Price increase from $0 to $100
Xbox Demand
6000000
4000000
3000000
2000000
1000000
New
Income
Xbox Demand
5000000
0
0
100 200 300 400 500 600 700 800 900
Xbox Price
85
2.2.3 Xbox 720- Example
Price increase from $500 to $600
Xbox Demand
6000000
Xbox Demand
5000000
4000000
3000000
Old Income
2000000
New
Income
1000000
0
0
100 200 300 400 500 600 700 800 900
Xbox Price
86
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) 87
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)
88
2.2.3 Elasticity Example 1
Let 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:
89
2.2.3 Elasticity Example 1
Let 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
90
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?
91
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.
92
-This has policy implications…
2.2.3 Elastic Xboxes
Let 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
93
2.2.3 Elastic Xboxes
Find 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)
94
2.2.3 Elastic Xboxes
From 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)
95
2.2.3 Making Microsoft Money
If 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.
96
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)
97
2.2.3 More Elasticity Exercises
Let q = 100-2p
1) Find Elasticities at p=5, 20 and 40
2) Formulate Economic Advice at these points
Let q = 200+2p-4p2
1) Find Elasticities at p=0, 5 and 10
2) Formulate Economic Advice at these points
98
2.2.3 Elastic Logs
ANOTHER 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)
99
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.
100
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)
1) Find Elasticities at p=0.5, 1, and 2
101