Differentiation
• Purpose- to determine instantaneous rate of
change
Eg: instantaneous rate of change in total cost per
unit of the good
We will learn
• Marginal Demand, Marginal Revenue, Marginal
Cost, and Marginal Profit
Marginal Cost : MC(q)
• What is Marginal cost?
The cost per unit at a given level of production
EX:
Recall Dinner problem
C(q) = C0 + VC(q).
= 9000+177*q0.633
MC(q)- the marginal cost at q dinners
MC(100)- gives us the marginal cost at 100 dinners
This means the cost per unit at 100 dinners
How to find MC(q) ? We will learn 3 plans
Marginal Analysis
• First Plan
• Cost of one more unit
• MC q  Cq  1  Cq
Marginal Analysis
• Ex. Suppose the cost for producing a particular
item is given by C q   8000  127q 0.607 where q
is quantity in whole units. Approximate
MC(500).
MC 500  C 501  C 500
•

 
 8000  127  5010.607  8000  127  500 0.607
 13528.47335  13521.77255
 $6.70 per unit

Marginal Analysis
• Second Plan
• Average cost of one more and one less unit
C q  1  C q  1
MC q  
•
2

C q  1  C q   C q   C q  1 

 derived from

2


Marginal Analysis
• Ex. Suppose the cost for producing a particular
item is given by C q   8000  127q 0.607 where q
is quantity in whole units. Approximate
MC(500).
C 501  C 499
MC 500 
•
2
8000  127  5010.607  8000  127  499 0.607

2
13528.47335  13515.06648

2
 $6.70 per unit

 

Marginal Analysis
•
Final Plan
•
Average cost of fractionally more and fractionally less units
difference quotients
•
C q  h   C q  h 
MC q   lim
h 0
2h
•
C q  h   C q  h 
Typically use MC q  
with h = 0.001
2h
Marginal Analysis
• Ex. Suppose the cost for producing a particular
item is given by C q   8000  127q 0.607 where q
is quantity in whole units. Approximate
MC(500).
C 500.001  C 499.999
MC 500 
•
2  0.001
8000  127  500.0010.607  8000  127  499.9990.607

0.002
13521.77926  13521.76585

0.002
 $6.71 per unit

In terms of money, the marginal cost at the
production level of 500, $6.71 per unit
 

Marginal Analysis
• Use “Final Plan” to determine answers
• All marginal functions defined similarly
Rq  h   Rq  h 


MR q  lim
h 0
•
2h
C q  h   C q  h 
MC q   lim
h 0
2h
Pq  h   Pq  h 
MP q   lim
h 0
2h
Marginal Analysis
• Graphs
Demand
Marginal Demand
D(q) is always decreasing
All the difference quotients for marginal demand are negative
MD(q) is always negative
Marginal
revenue 0
Maximum
revenue
• Graphs
Marginal Analysis
Revenue
Marginal Revenue
q1
q1
R(q) is increasing
• difference quotients for marginal revenue are positive
•MR(q) is positive
R(q) is decreasing
• difference quotients for marginal revenue are negative
•MR(q) is negative
Marginal Analysis
• Graphs
Cost
Marginal Cost
Profit
Marginal Profit
Derivatives-part 2
• Difference quotients
f x  h   f x  h 
m  f x   lim
•
h 0
2h
• Called the derivative of f(x)
Derivatives
• Ex1. Evaluate f 3 if f x   2 x  6

2
f 3 
 

 6  2 2.999  6
2  0.001
14.0055471  13.99445674

0.002
0.0110903561

0.002
 5.5452
3.001
Derivatives
• Differentiating.xls file
• Graphs both function and derivative
• Can evaluate function and derivative
Derivatives
• Differentiating.xls
Definition
Formula for f (x )
=
Computation
f (x )
f ' (x )
x
=
Plot Interval
a
b
Increment
h
0.000001
#VALUE!
Constants
s
t
u
v
w
DERIVATIVE
FUNCTION
1
1
0.8
0.8
0.6
0.6
f ' (x )
f (x )
0.4
0.4
0.2
0.2
0
0
0
0.2
0.4
0.6
x
0.8
1
0
0.2
0.4
0.6
x
0.8
1
Derivatives
• Use Differentiating.xls to graph the derivative of
f x   2 x  6 on the interval [-2, 8]. Then
DERIVATIVE
evaluate f 3
.
200
150
f ' (x )
100
50
f 3  5.5452
0
-5
0
5
x
10
f x   2  6
x
FUNCTION & DERIVATIVE
function
derivative
300
250
200
f(x)
f'(x)
150
100
50
0
-4
-2
0
2
4
x
6
8
10
Differentiating.xslm
Definition
Formula for f (x )
7
x
3
Computation
f (x )
f ' (x )
14
5.545
Plot Interval
a
b
-2
8
Increment
h
0.000001
Key points
1)Formula for the function in x
2)Plot Interval is ESSENTIAL
3)You can use the computation cells to evaluate the f(x) & f’(x)
at different values
4)If using Office 2007, save it as a macro enabled file
Derivatives
• Properties
If
If
f x   c (c is a constant) then
f x  mx(m is a constant) then
If g x  a  f x
f x   0
f x   m
then g x  a  f x
If hx  f x  g x then hx  f x  g x
Derivatives
• Tangent line approximations
• Useful for easy approximations to complicated
functions
• Need a point and a slope (a derivative)
• Use y = mx +b
Derivatives
• Ex. Determine the equation of the tangent line to
f x   2 x  6 at x = 3.
• Recall
f 3  5.5452
and we have the point (3, 14)
• Tangent line is y = 5.5452x – 2.6356
Derivatives
• Project (Marginal Revenue)
- Typically MRq  Rq
- In project,
MRq  1000  Rq
Rq  h   Rq  h 
million dollars
R q  
units are thousand units
2h
Recall:Revenue function-R(q)
• Revenue in million dollars R(q)
typically
MRq  Rq
1000000  Rq 
MR q  
1000
 1000 Rq 
• Why do this conversion?
Marginal Revenue in dollars per drive
24
Derivatives
• Project (Marginal Cost)



q
MC
q

C
- Typically
- In project, MCq  1000  C q
C q  h   C q  h 
million dollars
C q  
units are thousand units
2h
Derivatives
• Project (Marginal Cost)
- Marginal Cost is given in original data
- Cost per unit at different production levels
- Use IF function in Excel
Derivatives
• Project (Marginal Profit)
MP(q) = MR(q) – MC(q)
- If MP(q) > 0, profit is increasing
- If MR(q) > MC(q), profit is increasing
- If MP(q) < 0, profit is decreasing
- If MR(q) < MC(q), profit is decreasing
Derivatives
• Project (Marginal Cost)
- Calculate MC(q)
Nested If function, the if function using values for
Q1-4 & 6
- IF(q<=800,160,IF(q<=1200,128,72))
In the GOLDEN sheet need to use cell referencing
for IF function because we will make copies of
it, and do other project questions
=IF(B30<$E$20,$D$20,IF(B30<$E$22,$D$21,$D$22))
Recall -Production cost estimates
•
•
Fixed overhead cost - $ 135,000,000
Variable cost (Used for the MC(q)
function)
1) First 800,000 - $ 160 per drive
2) Next 400,000- $ 128 per drive
3) All drives after the first 1,200,000$ 72 per drive
Derivatives
• Project (Maximum Profit)
- Maximum profit occurs when MP(q) = 0
- Max profit occurs when MR(q) = MC(q)
- Estimate quantity from graph of Profit
- Estimate quantity from graph of Marginal
Profit
Derivatives-change
• Project (Answering Questions 1-3)
1. What price? $285.88
2. What quantity? 1262(K’s) units
3. What profit? $42.17 million
Derivatives
• Project (What to do)
- Create one graph showing MR and MC
- Create one graph showing MP
- Prepare computational cells answering your
team’s questions 1- 3
Marginal AnalysisR ( q  h)  R ( q  h)
2h
 MR(q) = R′(q) ∙ 1,000
 R(q) 


where h = 0.000001
0.160 if 0  q  800

C (q)  0.128 if 800  q  1, 200
0.072 if q  1, 200

160 if 0  q  800

MC (q)  C (q) 1, 000  128 if 800  q  1, 200
 72 if q  1, 200

Marketing Project
Marginal Analysis R(q) 
R ( q  h)  R ( q  h)
2h
where h = 0.000001
 In Excel we use derivative of R(q)
 R(q)=aq^3+bq^2+cq
 R’(q)=(a*3*q^2+b*2*q+c)/1000

Marketing Project
Marginal Analysis (continued)Marginal Revenue and Cost
$600
MR(q)
$ Per Drive
$400
MC(q)
$200
$0
0
400
800
1,200
1,600
2,000
2,400
2,800
-$200
-$400
-$600
q (K's)
Marketing Project
Marginal Analysis
 MP(q) = MR(q) – MC(q)
Marginal Profit
MP(q) $ Per Drive
$400
$200
$0
0
400
800
1,200
1,600
2,000
-$200
-$400
q (K's)
 We will use Solver to find the exact value of q for
which MP(q) = 0. Here we estimate from the graph
Marketing Project
Profit Function
 The profit function, P(q), gives the relationship
between the profit and quantity produced and sold.
 P(q) = R(q) – C(q)
P (q ) (M's)
Profit Function
$70
$60
$50
$40
$30
$20
$10
$0
-$10 0
-$20
400
800
1,200
q (K's)
1,600
2,000
Goals
•
1. What price should Card Tech put on the drives,
in order to achieve the maximum profit?
•
2. How many drives might they expect to sell at
the optimal price?
•
3. What maximum profit can be expected from
sales of the 12-GB?
•
4. How sensitive is profit to changes from the
optimal quantity of drives, as found in Question 2?
•
5. What is the consumer surplus if profit is
maximized?
38
Goals-Contd.
•
6. What profit could Card Tech expect, if they price the
drives at $299.99?
•
7. How much should Card Tech pay for an advertising
campaign that would increase demand for the 12-GB drives by 10%
at all price levels?
•
8. How would the 10% increase in demand effect the
optimal price of the drives?
•
9. Would it be wise for Card Tech to put $15,000,000 into
training and streamlining which would reduce the variable production
costs by 7% for the coming year?
39