1 2. Economic Applications of Single- Variable Calculus Derivative Origins Single Variable...

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2. Economic Applications of Single-Variable Calculus

• Derivative Origins• Single Variable Derivatives• Economic uses of Derivatives

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2.1 Derivatives of Single-Variable Functions

2.2 Applications using Derivatives

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In economics, derivatives are used in various ways:

Marginal amounts (slope) Maximization Minimization Graphing Elasticity and Growth

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Slope:

-consider the following graph of demand

-what is the impact of a $55 drop in price?

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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).

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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?

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--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

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2.1 – Tangents

The green tangent line represents the instantaneous slope

Slopes

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Price

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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 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

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Function:

Each X

Has 1 Y

y=0+2sin(2pi*x/14)+2cos(2pi*x/14)

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-1

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4

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x

y 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) = 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

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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.

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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

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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:

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

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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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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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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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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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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

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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)

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-1

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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

)

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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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4) Exponents

If y = bx

y’ = bxln(b)

Therefore

If y = ex

y’ = ex

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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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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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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

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x=15+10t-t*t

0

10

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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

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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’’

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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

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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-yd(14)/dx=d(7x)/dx+d(9x2)/dx-dy/dx0 = 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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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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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.

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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

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1 2 3 4 5 6 7 8 9 10

Linear, Quadratic, and Lin-Log Graphs

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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

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-5

0

5

10

1 2 3 4 5 6 7 8

t

x x

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2.2.1 – Sample Graphs

x’’=-2, slope is decreasing; graph is concave

x=15+10t-t*t

0

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t

x x

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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

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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

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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

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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)

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t

x x

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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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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 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

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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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y=(x-5)(x-5)-3

-5

0

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25

1 2 3 4 5 6 7 8 9 10

x

y

(0,22)

(3.3,0) (6.7,0)

(5,-3)

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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

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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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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

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0

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50

60

70

-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.)

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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

0

10

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30

40

50

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t

x x

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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?

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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)

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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 +-

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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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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

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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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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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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

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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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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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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

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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

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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.

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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)

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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:

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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

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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)

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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)

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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



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)



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


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


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

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