Applied Calculus for Business Worksheets
By O. Pauline Chow
HACC
1
Math 110 Applied Calculus for Business
Lecture notes on Chapter 4.7 (audio dated 9/29/07 at 12:53 pm, 43 minutes long)
Relative and Percentage Rates of Changes
Elasticity
Elasticity of Demand
General Derivative Rules
1.
2.
3.
n −1
d 
n
( f ( x) )  = n ( f ( x) ) f '( x)
dx 
d
e f ( x )  = e f ( x ) f '( x)
dx 
d
1
f '( x)
ln ( f ( x) )  =
f '( x) =
dx 
f ( x)
f ( x)
Relative and Percentage Rates of Changes
Relative rate of change of a function f(x) is the ratio of the derivative to the function,
Percentage rate of change of a function f(x) is
f '( x)
(will be given on the test).
f ( x)
f '( x)
x100 .
f ( x)
Elasticity of demand
To study the effect of changes in price on demand, we will take a look at the elasticity of demand.
There is a general belief that increase in price will result in lower demand and decrease in price will result in higher demand.
Elasticity of demand, E, of a function is defined as the negative rate of change of the relative rate of change of the demand function
with respect to the relative rate of change of price function, E is a function of p.
Let x = f(p), where x is the number of items sold/produced and p is the price per unit.
Definition: E ( p ) = −
relative rate of change of demand x
relative rate of change of price p
f '( p)
pf '( p )
f ( p)
E ( p) = −
=−
1
f ( p)
p
E(p)
0 < E(p) < 1
Demand
Inelastic
E(p) > 1
Elastic
E(p) = 1
Unit
(will be given on the test)
Interpretation
Demand is not sensitive to changes in price. A change in price
produces a smaller changes in demand.
Demand is sensitive to changes in price. A change in price
produces a larger change in demand.
A change in price produces the same change in demand.
Revenue R(p) and Elasticity of Demand E(p)
R(p) = xp = pf(p)
R’(p) = p’ f(p) + p f’(p)
= f(p) + p f ’(p)
Applied Calculus for Business Worksheets
By O. Pauline Chow
HACC
 pf '( p ) 
= f ( p ) 1 +
f ( p ) 

R’(p) = f(p) [1 – E(p)] (will be given on the test)
Demand is inelastic if R’(p) > 0 and E(p) < 1. As price increases, revenue increases and vice versa.
Demand is elastic if R’(p) < 0 and E(p) > 1. As price increases, revenue decreases and vice versa.
You have to know above relationship on the test.
Find the relative rate of change of f(x).
Ex 1: f(x) = 50x – 0.01x2
f '( x)
50 − 0.02 x
=
f ( x) 50 x − 0.01x 2
Ex 2: f(x) = 15x + 2x ln x
1

15 +  2 ln x + 2 x 
f '( x)
17 + 2 ln x
x

=
=
f ( x)
15 x + 2 x ln x
15 x + 2 x ln x
Ex 3: Given: x = f(p) = 875 – p – 0.05p2. Find E(50), E(70), E(100). Comment elasticity.
f ’(p) = –1 – 0.10p
p [ −1 − 0.10 p ]
pf '( p )
p + 0.10 p 2
E(p) = −
=−
=
2
f ( p)
875 − p − 0.05 p
875 − p − 0.05 p 2
E(50) = =
(50) + 0.10(50) 2
300 3
=
= < 1 , demand is inelastic.
875 − (50) − 0.05(50) 2 700 7
E(70) = =
(70) + 0.10(70) 2
560
=
= 1 , demand is unit.
875 − (70) − 0.05(70) 2 560
E(100) =
(100) + 0.10(100)2
1100
=
= 4 > 1 , demand is elastic.
2
275
875 − (100) − 0.05(100)
Ex 4: Given: x = f(p) = 10(p – 9)2 . Find the values of p for which the demand is elastic and the values for which the demand is
inelastic.
First you need to find E(p).
f ’(p) = 20(p – 9)
E ( p) = −
−2 p
pf '( p)
20 p[ p − 9]
=−
=
f ( p)
10( p − 9)2 ( p − 9)
Next you need to find the values of p when the demand is elastic.
E ( p) > 1
demand is elastic
−2 p
>1
p −9
−2 p
−1 > 0
p −9
2
Applied Calculus for Business Worksheets
By O. Pauline Chow
HACC
−2 p p − 9
−
>0
p −9 p −9
−2 p − p + 9
>0
p−9
−3 p + 9
>0
p −9
Now use the sign graph of
−3 p + 9
to solve the inequality.
p −9
p – 9 = 0, p = 9
–3p + 9 = 0, p = 3
__– – – |_+ + +|_– – – _
3
9
So, if E(p) > 1, 3 < p < 9. Demand is elastic when 3 < p < 9. Revenue is decreasing when price increases.
If 0 < E(p) < 1, the demand is inelastic when 0 < p < 3 and revenue is increasing when price increases.
Note: For graphing calculator user: you may use the TI to solve the inequality:
E ( p) > 1
−2 p
>1
p −9
demand is elastic
Enter y1 = –2x/(x – 9) and y2 = 1
Since
−2 p
> 1 means the graph of y1 = –2x/(x – 9) is greater than y2 = 1, x is between 3 and 9. See the graph below:
p −9
Ex 5: Given: x = f ( p) = 30 − 5 p . Find the values of p for which the demand is elastic and the values for which the demand is
inelastic.
First you need to find E(p).
5
1

f '( p ) = −5  p −1/ 2  = −
2 p
2

3
Applied Calculus for Business Worksheets
By O. Pauline Chow
HACC
E ( p) = −
4

5 
p −

 2 p 
5 p
p
pf '( p )
=−
=
=
f ( p)
30 − 5 p
2(30 − 5 p ) 2(6 − p )
Next you need to find the values of p when the demand is elastic.
E ( p) > 1
demand is elastic
p
>1
2(6 − p )
p
2(6 − p )
p
2(6 − p )
−1 > 0
−
2(6 − p )
2(6 − p )
p − 12 + 2 p
2(6 − p )
3 p − 12
2(6 − p )
3( p − 4)
2(6 − p )
>0
>0
>0
>0
Now use the sign graph of
3( p − 4)
2(6 − p )
to solve the inequality.
p −4 = 0
p =4
p = 16
6− p
6=
p
p = 36
_–_– – |_+ + +|_– – – _
16
36
So, if E(p) > 1, 16 < p <36. Demand is elastic when 16 < p < 36. Revenue is decreasing when the price increases.
So, if 0 < E(p) < 1, 0 < p < 16. Demand is inelastic. Revenue is increasing when the price increases.
TI user:
E ( p) > 1
p
2(6 − p )
demand is elastic
>1
Enter y1 = x1/2/(12 – 2p1/2) and y2 = 1
Applied Calculus for Business Worksheets
By O. Pauline Chow
HACC
5
So, demand is elastic when 16 < p < 36 and inelastic when 0 < p < 16.
Ex 6: The fast food restaurant can produce an order of fries for $0.40. If the restaurant’s daily sales are increasing at the rate of 15 per
day, how fast is their daily cost for fries increasing?
cost per order of fries = 0.40 and rate of change of demand is 15 per day
The daily cost is cost per order of fries times the demand per day = 0.40(15) = $6 per day
Ex 7: The price-demand equation for hamburgers at a fast food restaurant is x + 400p = 2000. Currently, the price of a hamburger is
$3.00, will a 10% price increase cause the revenue to increase or decrease? [Hint: Find E(p)]
x = 2000 – 400p
E ( p) = −
pf '( p)
p[−400]
400 p
p
=−
=
=
f ( p)
2000 − 400 p 400(5 − p ) 5 − p
p = 3, E (3) =
3
= 1.5 > 1
5−3
If E(3) > 1, then demand is elastic.
R’(p) = f(p)[1 – E(p)] < 0; R decreases
So, 10% price increase will cause a decrease in revenue.