Download Business and Economic Applications

Business and
Economic
Applications
Summary of Business
Terms and Formulas
 x is the number of units produced (or sold)
 p is the price per unit
 R is the total revenue from selling x units
R = xp
 C is the total cost of producing x units
C is the average cost per unit
C
C
x
Summary of Business
Terms and Formulas
 P is the total profit from selling x units
P=R–C
 The break-even point is the number of
units for which R = C.
Marginals
dR
 (Marginal revenue)  (extra revenue
dx
from selling one additional unit)
dC
 (Marginal cost)  (extra cost of producing
dx
one additional unit)
dP
 (Marginal profit)  (extra profit from selling
dx
one additional unit)
Using Marginals as
Approximations
 A manufacturer determines that the profit
derived from selling x units of a certain item is
given by P = 0.0002x3 + 10x.
 a. Find the marginal profit for a production level
of 50 units
 b. Compare this with the actual gain in profit
obtained by increasing the production from 50
to 51 units.
Demand Function
 The number of units x that consumers
are willing to purchase at a given price p
is defined as the demand function
 p = f(x)
Finding the Demand
Function
 A business sells 2000 items per month at
a price of $10 each. It is predicted that
monthly sales will increase by 250 items
for each $0.25 reduction in price. Find
the demand function corresponding to
this prediction.
Steps
 First find the number of units produced:
 10  p 
x  2000  250 

 0.25 
 2000  1000(10  p )
 2000  10000  1000 p
 12, 000  1000 p
Steps
 Now solve this equation for p
x  12, 000  1000 p
x
x
 12  p or 12 
,
1000
1000
x  2000
Finding Marginal Revenue
 A fast-food restaurant has determine that
the monthly demand for its hamburgers is
60, 000  x
p
20, 000
 Find the increase in revenue per hamburger
(marginal revenue) for monthly sales of
20,000 hamburgers.
Finding Marginal Revenue
 Because the total revenue is given by
 R = xp, you have
1
 60, 000  x 
2
R  xp  x 

(60,
000
x

x
)

 20, 000  20, 000
 and the marginal revenue is
dR
1

(60, 000  2 x)
dx 20, 000
Finding Marginal Revenue
 When x = 20,000 the marginal revenue
dR
1
(20, 000) 
60, 000  2(20, 000)
dx
20, 000
20, 000

 $1/ unit
20, 000
 Now lets look at the graph. Notice that
as the price decreases, more
hamburgers are sold. (Make sense?)
Finding Marginal Profit
 Suppose that the cost of producing those
same x hamburgers is
 C = 5000 + 0.56x
 (Fixed costs = $5000; variable costs are
56¢ per hamburger)
Finding Marginal Profit
 Find the total profit and the marginal
profit for 20,000, for 24,400, and for
30,000 units.
 Solution: Because P = R – C, you can
use the revenue function to obtain
Finding Marginal Profit
1
P
(60, 000 x  x 2 )  (5000  0.56 x)
20, 000
60, 000 x
x2


 5000  0.56 x
20, 000 20, 000
x2
 3x  0.56 x 
 5000
20, 000
x2
 2.44 x 
 5000
20, 000
Finding Marginal Profit
 Now do the derivative to find a marginal
dP
x
 2.44 
dx
10, 000
Demand 20,000
Profit
24,400
30,000
$23,800 $24,768 $23,200
Marginal $0.44
profit
$0.00
- $0.56
Finding Maximum Profit
 In marketing a certain item, a business
has discovered that the demand for the
item is
50
p
x
 The cost of producing x items is given by
C = 0.5x + 500. Find the price per unit
that yields a maximum profit.
Finding Maximum Profit
 From the given cost function, you obtain
 P = R – C = xp – (0.5x + 500).
 Substituting for p (from the demand function)
produces
 50 
P  x
  (0.5 x  500)  50 x  0.5 x  500
 x
Finding Maximum Profit
 To find maximums using calculus, you now
do a derivative of the profit equation and
then set it equal to zero.
dP 25

 0.5
dx
x
25
0
 0.5
x
25
0.5 

x
x  50
0.5 x  25
x  2500
Finding Maximum Profit
 This gives us the number of units needed to
be produced in order to get the maximum
profit. How do we find the price?
50
50
p

 $1.00
2500 50
Minimizing the Average
Cost
 A company estimates that the cost (in
dollars) of producing x units of a certain
product is given by
 C = 800 + 0.4x + 0.0002x2.
 Find the production level that minimizes
the average cost per unit.
Minimizing Average Cost
 Substituting from the given equation for C
produces
C 800  0.04 x  0.0002 x 2
C 
x
x
800

 0.04  0.0002 x
x
Finding the derivative and setting it equal to 0 yields
dC
800
  2  0.0002  0
dx
x
800
0.0002  2
0.0002 x 2  800
x
800
2
x 
x 2  4, 000, 000  x  2000 units
0.0002
Practice
 Your turn
 Homework (From CD Appendix G)
 p. G5 problems 1 – 33 odd