Copyright © 2016, 2012 Pearson Education, Inc. 2.5 - 1.

Copyright © 2016, 2012 Pearson Education, Inc. 2.5 - 1

2.5


Maximum-Minimum Problems; Business and Economics Applications

OBJECTIVE• Solve maximum and minimum problems using calculus.


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.

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 – 2.4 to determine the maximum or minimum values and the points at which they occur.



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.



Example 1 (continued):2nd write an equation for the volume of the box.

Note that x must be between 0 and 4. So, we need to maximize the volume equation on the interval (0, 4).


V

V x

l w h (8 2 ) (8 2 )x x x

2(64 32 4 )x x x 3 24 32 64x x x

V x

V x


Example 1 (continued):

is the only critical value in (0, 4). So, we can use the second derivative.


V 212 64 64x x 23 16 16x x

(3 4)( 4)x x

0

3 4 0x 4 0or x

4

3x 4or x

0

0


Example 1 (concluded):

Thus, the volume is maximized when the square corners are inches. The maximum volume is


V x

4

3V

24 64x

424 64

3

32 0

4

3V

4

3V

3 24 4 4

4 32 643 3 3

32537 in

27

4

3V



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) of liquid? (Hint: and ).

Going from Example 1, we can visualize that the dimensions of the box will be

31 L=1000 cm 3 31 in 16.38 cm

8.5 2 ,11 2 ,x x x



Quick Check 1 Continued

Next lets find the equation for volume:

Since, both and , and .

So, must be between 0 and -4.25. So we must maximize the volume equation on the interval .

V l w h (11 2 )(8.5 2 )V x x x

3 24 39 93.5V x x x

x(0,4.25)

11 2 0x 8.5 2 0x 0 5.5x 0 4.5x




Using the quadratic equation:

Since does not fall into the range, .

212 78 93.5V x x 0

2 4

2

b b acx

a

2( 78) ( 78) 4(12)(93.5)

2(12)x

78 2 399

24x

1.585x 4.917x

4.917 1.585x

or,



Quick Check 1 Concluded

Therefore there is a maximum at

Thus the dimensions are approximately 1.585 in. by 5.33 in. by 7.83 in., giving us a volume of or , which is greater than 1 Liter. Thus the box can hold 1 Liter of water.

24 78V x

1.585 24 1.585 78V 1.585 38.04 78V 1.585 39.96 0V

1.585.x

366.15 in. 31083.5 cm.


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 The manufactureralso determines that the total cost of producing x units is given by

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?

( ) 1000 .p x x

( ) 3000 2 .C x x



Example 2 (continued): a)

b)


Revenue quantity price( )R x x p

(1000 )x x21000x x

( )R x

( )R x

Profit Total Revenue Total Cost( )P x R x C x

21000 3000 20x x x 2 980 3000x x

( )P x

( )P x


Example 2 (continued): c)

Since there is only one critical value, we can use the second derivative to determine whether or not it yields a maximum or minimum.

Since P (x) is negative, x = 490 yields a maximum. Thus, profit is maximized when 490 units are bought and sold.


( )P x 2 980x 2xx

0

980490


Example 2 (concluded): d) The maximum profit is given by

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


(490)P 2(490) 980(490) 3000 $237,100.(490)P

(490)p 1000 490$510.(490)p



Quick Check 2

Repeat Example 2 with the price function and the cost function . Round your answers when necessary.

a.) Find the total revenue

( ) 1750 2p x x ( ) 2250 15C x x

( )R x

2

Revenue quantity price

( )

( ) (1750 2 )

( ) 1750 2

R x x p

R x x x

R x x x




b.) Find the total Profit:

c.) Find the number of units to produce by using the First Derivative Test to find a critical value:

We can use the second derivative to see if this is a maximum or minimum. Since , we know profits are maximized when 434 units are sold.

2

2

Profit Total Revenue Total Cost

( )

1750 2 2250 15

( ) 2 1735 2250

P x R x C x

P(x) x x x

P x x x

( ) 4 1735 0,P x x 433.75x

( ) 4 0P x



Quick Check 2 concluded

d.) Find the Maximum Profit:

The maximum profit is given by:

Thus the maximum profit is $374,028.

e.) Find the price per unit.

The price per unit to achieve this maximum profit is:

2(434) 2(434) 1735(434) 2250P

(434) 374,028P

(434) 1750 2(434)p , (434) 1750 868p , (434) $882p


THEOREM 10

Maximum profit occurs at those x-values for which

R (x) = C (x) and R (x) < C (x).



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?



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

# of people ticket price # of people 4R x

(1000 50 )(26 ) (1000 50 ) 4R x x x x

226,000 1000 1300 50 4000 200R x x x x x

250 500 30,000R x x x


Example 3 (continued): To maximize R(x), we find R (x) and solve for critical values.

Since there is only one critical value, we can use the second derivative to determine if it yields a maximum or minimum.


( )R x 100 500x 100x

x

0

5005


Example 3 (concluded):

Thus, x = 5 yields a maximum revenue. So, the theater should charge

$26 – $5 = $21 per ticket.


( )R x 100(5)R 100



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:

( ) (20,000 800 )(30 ) (20,000 800 ) 8R x x x x 2( ) 600,000 4,000 800 160,000 6,400R x x x x

2( ) 800 10,400 760,000R x x x



Quick Check 3 Concluded

Next we find :

Then find where :

To check to make sure this is a maximum, we need to use the second derivative. Since we know that this is in fact a maximum of the revenue equation.

Substitute x back into the formula for ticket price we get:

( )R x ( ) 1,600 10,400R x x

( ) 0R x ( ) 1,600 10,400 0R x x

1,600 10,400x 6.5x

( ) 1,600 0,R x

$30 $6.50 $23.50



Section Summary

• In many real-life applications, we need to determine the minimum or maximum value of some function modeling a situation.

• Maximum profit occurs at those x-values for which R (x) = C (x) and R (x) < C (x).

Copyright © 2016, 2012 Pearson Education, Inc. 2.5 - 1.

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