Maximum-Minimum Problems; Business
and Economics Applications
2.5
OBJECTIVE
• Solve maximum and minimum problems
using calculus.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
A Strategy for Solving Maximum-Minimum
Problems:
1. Read the problem carefully. If relevant, make a
drawing.
2. Make a list of appropriate variables and constants,
noting what varies, what stays fixed, and what units
are used. Label the measurements on your drawing,
if one exists.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
A Strategy for Solving Maximum-Minimum
Problems (concluded):
3. Translate the problem to an equation involving a
quantity Q to be maximized or minimized. Try to
represent Q in terms of the variables of step (2).
4.Try to express Q as a function of one variable. Use
the procedures developed in sections 2.1 – 2.3 to
determine the maximum or minimum values and the
points at which they occur.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 1: From a thin piece of cardboard 8 in. by 8
in., square corners are cut out so that the sides can be
folded up to make a box. What dimensions will yield a
box of maximum volume? What is the maximum
volume?
1st make a drawing in which x is the length of each
square to be cut.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 1 (continued):
2nd write an equation for the volume of the box.

V
V  x
l wh
 (8  2 x)  (8  2 x)  x
V  x   (64  32 x  4 x 2 )  x
3
2

4 x  32 x  64 x
V  x
Note that x must be between 0 and 4. So, we need to
maximize the volume equation on the interval (0, 4).
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 1 (continued):
2

12
x
 64 x  64

V
3x 2  16 x  16
(3x  4)( x  4)

0


0
0
3x  4  0 or x  4  0
4
or
x4
x
3
4
is the only critical value in (0, 4). So, we can use
3
the second derivative.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 1 (concluded):
24 x  64
V   x  
4
V   
3
4

V   
3

4
24    64
3

32

0
Thus, the volume is maximized when the square
4
corners are inches. The maximum volume is
3
4
V 
3
4
V 
3
3
2
4
4
4
 4  3   32  3   64  3 
 
 
 
25 3
 37 27 in
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 1
Repeat Example 1 with a sheet of cardboard measuring 8.5 in. by 11
in. (the size of a typical sheet of paper). Will this box hold 1 liter (L)
3
3
3
of liquid? (Hint: 1 L=1000 cm and 1 in  16.38 cm ).
Going from Example 1, we can visualize that the dimensions of the
box will be 8.5  2 x,11  2 x, x
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 1 Continued
Next lets find the equation for volume:
V  l  wh
V  (11  2 x)(8.5  2 x) x
V  4 x3  39 x 2  93.5 x
Since, both 11  2 x  0 and 8.5  2 x  0 , 0  x  5.5 and 0  x  4.25.
So, x must be between 0 and 4.25. So we must maximize the
volume equation on the interval (0,4.25) .
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 1 Continued
V   12 x 2  78 x  93.5  0
b  b 2  4ac
Using the quadratic equation: x 
2a
(78)  (78) 2  4(12)(93.5)
x
2(12)
78  2 399
x
, x  4.917 or x  1.585
24
Since 4.917 does not fall into the range, x  1.585 .
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 1 Concluded
V   24 x  78
V  1.585  24 1.585  78
V  1.585  38.04  78
V  1.585  39.96  0
Therefore there is a maximum at x  1.585.
Thus the dimensions are approximately 1.585 in. by 5.33 in. by 7.83
in., giving us a volume of 66.15 in.3or 1083.5 cm.3, which is greater
than 1 Liter. Thus the box can hold 1 Liter of water.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 2: A stereo manufacturer determines that in
order to sell x units of a new stereo, the price per unit,
in dollars, must be p( x)  1000  x. The manufacturer
also determines that the total cost of producing x units
is given by C ( x)  3000  2 x.
a) Find the total revenue R(x).
b) Find the total profit P(x).
c) How many units must the company produce and
sell in order to maximize profit?
d) What is the maximum profit?
e) What price per unit must be charged in order to
make this maximum profit?
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 2 (continued):
a) Revenue  quantity  price

R( x)
x p
R( x)

x(1000  x)

R( x)
1000x  x 2
b) Profit
P( x)
P( x)
P( x)
 Total Revenue  Total Cost
 R x  C  x
 1000 x  x 2   3000  20 x 
2
  x  980 x  3000
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 2 (continued):
c) P( x)  2 x  980
2x

0
 980
x
 490
Since there is only one critical value, we can use the
second derivative to determine whether or not it yields
a maximum or minimum.
P(x)  2
Since P(x) is negative, x = 490 yields a maximum.
Thus, profit is maximized when 490 units are bought
and sold.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 2 (concluded):
d) The maximum profit is given by
2
P (490)  (490)  980(490)  3000
P (490) 
$237,100.
Thus, the stereo manufacturer makes a maximum
profit of $237,100 when 490 units are bought and sold.
e) The price per unit to achieve this maximum profit is
p (490)  1000  490
$510.
p (490) 
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 2
Repeat Example 2 with the price function p ( x)  1750  2 x and
the cost function C ( x)  2250  15 x . Round your answers when
necessary.
a.) Find the total revenue R ( x )
Revenue  quantity  price
R( x)

x p
R( x)

x(1750  2 x)
R( x)

1750 x  2 x 2
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 2 Continued
b.) Find the total Profit: Profit 
P( x) 
Total Revenue  Total Cost
P(x)

1750 x  2 x 2   2250  15 x 
P( x)
 2 x 2  1735 x  2250
R  x  C  x
c.) Find the number of units to produce by using the First Derivative
Test to find a critical value: P( x)  4 x  1735  0, x  433.75
We can use the second derivative to see if this is a maximum or
minimum. Since P( x)  4  0 , we know profits are maximized
when 434 units are sold.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 2 concluded
d.) Find the Maximum Profit:
The maximum profit is given by:
P(434)  2(434)2  1735(434)  2250
P(434)  374,028
Thus the maximum profit is $374,028.
e.) Find the price per unit.
The price per unit to achieve this maximum profit is:
p(434)  1750  2(434) , p(434)  1750  868 , p(434)  $882
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
THEOREM 10
Maximum profit occurs at those x-values for which
R(x) = C(x) and R(x) < C(x).
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 3: Promoters of international fund-raising
concerts must walk a fine line between profit and loss,
especially when determining the price to charge for
admission to closed-circuit TV showings in local
theaters. By keeping records, a theater determines
that, at an admission price of $26, it averages 1000
people in attendance. For every drop in price of $1, it
gains 50 customers. Each customer spends an
average of $4 on concessions. What admission price
should the theater charge in order to maximize total
revenue?
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 3 (continued):
Let x = the number of dollars by which the price of
$26 should be decreased (if x is negative, the price
should be increased).
Revenue  Rev. from tickets  Rev. from concessions
R  x   # of people  ticket price  # of people  4
R  x   (1000  50 x)(26  x)  (1000  50 x)  4
R  x   26,000  1000 x  1300 x  50 x 2  4000  200 x
R  x   50 x 2  500 x  30,000
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 3 (continued):
To maximize R(x), we find R(x) and solve for critical
values.
0
R( x)  100 x  500 
100x  500
x  5
Since there is only one critical value, we can use the
second derivative to determine if it yields a maximum
or minimum.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Example 3 (concluded):
R( x)  100
R(5)  100
Thus, x = 5 yields a maximum revenue. So, the theater
should charge
$26 – $5 = $21 per ticket.
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 3
A baseball team charges $30 per ticket and averages 20,000 people in
attendance per game. Each person spends an average of $8 on
concessions. For every drop of $1 in the ticket price, the attendance
rises by 800 people. What ticket price should the team charge to
maximize total revenue?
From Example 3, we can use the Revenue formula to find the ticket
price to maximize income:
R( x)  (20,000  800 x)(30  x)  (20,000  800 x)  8
R( x)  600,000  4,000 x  800 x 2  160,000  6,400 x
R( x)  800 x 2  10,400 x  760,000
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Quick Check 3 Concluded
Next we find R( x) : R( x)  1,600 x  10, 400
Then find where R( x)  0 :
R( x)  1,600 x  10, 400  0
1,600 x  10,400
x  6.5
To check to make sure this is a maximum, we need to use the
second derivative. Since R( x)  1,600  0, we know that this is
in fact a maximum of the revenue equation.
Substitute x back into the formula for ticket price we get:
$30  $6.50  $23.50
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2.5 Maximum-Minimum Problems; Business and
Economics Applications
Section Summary
• In many real-life applications, we wish to determine the minimum
or maximum value of some function modeling a situation.
• Identify a realistic interval for the domain of the input variable. If it
is closed interval, its endpoints should be considered as possible
critical values.
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