Assignments One
P12
61, 62, 64
61. IMMUNIZATION Suppose that during a nationwide program to
immunize the population against a certain form of influenza, public
health officials found that the cost of inoculating x% of the population
was approximately C ( x) 
150 x
million dollars.
200  x
a. What is the domain of the function C?
b. For what values of x does C(x) have a practical interpretation in this
context?
c. What was the cost of inoculating the first 50% of the population?
d. What was the cost of inoculating the second 50% of the population?
e. What percentage of the population had been inoculated by the time
37.5 million dollars had been spent?
Solution
a. Since division by any number other than 0 is possible, the domain of C
is the set of all real numbers x  200 .
b. C(x) has a practical interpretation for 0  x  100 since x means the
number of percent of the population.
c. The cost of inoculating the first 50% of the population is the value of
total cost function when x  50 . That is
C (50)  C (0) 
150  50
 $50 million
200  50
d. The cost of inoculating the second 50% of the population is the
difference between the value of total cost function when x  100 and the
value when x  50 . That is
150 100
 50  $100 million
200  100
e. Set C (x) equal to 37.5 and solve for x to get
C (100)  C (50) 
150 x
 37.5
200  x
x  40
That is, 40% of the population had been inoculated by the time 37.5
million dollars had been spent.
62. POSITION OF A MOVING OBJECT A ball has been dropped
from the top of a building. Its height (in feet) after t seconds is given by
the function H (t )  16t 2  256 .
a. What is the height of the ball after 2 seconds?
b. How far will the ball travel during the third second?
c. How tall is the building?
d. When will the ball hit the ground?
Solution
a. After 2 seconds the height of the ball is
H (2)  16(2) 2  256  192 feet
b. During the third second the distance that the ball will travel is the
difference between the height of the ball after 2 second and the height
after 3 second. That is
H (2)  H (3)  192  (16(3) 2  256)  80 feet
c. The height of the building is the height of the ball when t  0 . That is
H (0)  16(0) 2  256  256 feet
d. Set H (t ) equal to 0 and solve for t to get
 16t 2  256  0
t 2  16
t   16  4
The time can not be negative, so we ignore –4.
That is, the ball will hit the ground after 4 seconds.
64. MANUFACTURING COST At a certain factory, the total cost of
manufacturing q units during the daily production run is
C (q)  q 2  q  900 dollars. On a typical workday, q (t )  25t units are
manufactured during the first t hours of a production run.
a. Express the total manufacturing cost as a function of t .
b. How much will have been spent on production by the end of the third
hour?
c. When will the total manufacturing cost reach $ 11,000?
Solution
a. Since the total cost is related to the variable q by the equation
C (q)  q 2  q  900
and the variable q is related to the variable t by the equation
q (t )  25t
it follows that the composition function
C (q(t ))  (25t ) 2  25t  900  625t 2  25t  900
C(q(t)) expresses total manufacturing cost as a function of t .
b. When
t 3 ,
C (q(3))  625  32  25  3  900  6600
That is, $6600 will have been spent on production by the end of third
hour.
c. Set C (q(t )) equal to 11,000 and solve for t to get
625t 2  25t  900  11000
t
 1  1  25  4  404  1  201

50
50
Since t  0 , we get t  4 . That is, the total manufacturing cost reach
$11,000 by the end of the forth hour.
P24
33
33. PROFIT Suppose that when the price of a certain commodity is p
dollars per unit, then x hundred units will be purchased by consumers,
where p  0.05 x  38 . The cost of producing x hundred units is
C ( x)  0.02 x 2  3 x  574.77 hundred dollars.
a. Express the profit P obtained from the sale of x hundred units as a
function of x. Sketch the graph of the profit function.
b. Use the profit curve found in part (a) to determine the level of
production x that results in maximum profit. What unit price p
corresponds to the maximum profit?
Solution
a. The profit function is
P( x)  R( x)  C ( x)  xp  C ( x)
 x(0.05 x  38)  (0.02 x 2  3x  574.77)
 0.07 x 2  35 x  574.77
which is a parabola that opens downward (Since A=-0.07<0) and has its
high point (vertex) at
x
B
 35

 250
2 A 2(0.07)
Thus, revenue is maximized when x=250 hundred units are produced,
and the corresponding maximum profit is P(250)=3800.23 hundred
dollars.
(250, 3800.23)
4,000
2,000
0
250
500
b. The profit is maximized when x  250 hundred units. And the
corresponding unit price is
p(250)  0.05(250)  38  $25.5
P37
20 , 28 , 34
P39
51
In Problems 20, 28 and 34, write an equation for the line with the given
properties
20.Through (-1, 2) with slope 2/3.
Solution
Use the formula y  y0  m( x  x0 ) with ( x0 , y0 )  (1,2) and m  2 / 3 to get
2
( x  1)
3
2
8
y  x
3
3
y2
Or
28. Through (-2,3) and (0,5).
The slope is
35
1
20
Use the point-slope formula with ( x0 , y0 )  (1,2) and m  1 to get
m
y 5 x
y  x5
34. Through (-1/2,1) and perpendicular to the line 2 x  5 y  3
Solution
By rewriting the equation 2 x  5 y  3 in the slope-intercept form
2
3
2
x  , we see that the line has slope m   . Then the line
5
5
5
1 5
perpendicular to this line must have the slope m'    , since the
m 2
y
required line contains (-1/2,1), we have
5
1
(x  )
2
2
5
9
y  x
2
4
y 1 
51. COLLEGE ADMISSIONS The average scores of incoming students
at an eastern liberal arts college in the SAT mathematics examination
have been declining at a constant rate in recent years. In 1995, the
average SAT score was 575, while in 2000 it was 545.
a. Express the average SAT score as a function of time.
b. If the trend continues, what will the average SAT score of incoming
students be in 2005?
c. If the trend continues, when will the average SAT score be 527?
Solution
a. Let t denote the number of year and y the average SAT score. Since y
changes at a constant rate with respect to t, the function relating y to x
must be linear. The line is through the points (1995, 575) and (2000, 545),
so the slope of the line will be
m
545  575
 6
2000  1995
y  6(t  1995)  575
To get
b. If the trend continues, in 2005, the average SAT score of incoming
students will be
f (2005)  6(2005  1995)  575  515
c. Set y equal to 527 and solve for t to get
 6(t  1995)  575  527
t  2003
That is, the average SAT score will be 527 in 2003 if the trend
continues.
P56
45 , 48
45. SUPPLY AND DEMAND Producers will supply x units of a
certain commodity to the market when the price is p  S (x) dollars per
unit, and consumers will demand (buy) x units when the price is
p  D(x) dollars per unit, where
S ( x)  2 x  15
and
D( x) 
385
x 1
a. Find the equilibrium profuction level xe and the equilibrium price
pe .
b.
Draw the supply and demand curves on the same graph.
c.
Where does the supply curve cross the y axis? Describe the
economic significance of this point.
Solution
a. Maket equilibrium occurs when
S ( x)  D( x)
385
2 x  15 
x 1
2
2 x  17 x  15  385
2 x 2  17 x  370  0
( x  10)( 2 x  37)  0
x  10
or x  18.5
Since only positive values of the production level x are meaningful,
we reject x  18.5 and conclude that equilibrium occurs when xe  10 .
The corresponding equilibrium price can be substiduting x  10 into
either the supply function or the demand function. Thus,
pe  S (10)  2(10)  15  35
b.
Equilibrium
point
(10, 35)
c. When the supply curve crosses the y axis,
S (0)  15
thus the supply curve crosses the y axis at (0,15) . This point mean
that no units will be produced until the price is at least $15.
48. BREAK-EVEN ANALYSIS A furniture manufacturer can selll
dining room tables for $70 apiece. The manufacturer’s total cost consists
of a fixed overhead of $8,000 plus production costs of 30 per table.
a. How many tables must the manufacturer sell to break even?
b. How many tables must the manufacturer sell to make a profit of
$6,000?
c.
What will be the manufacturer’s profit or loss if 150 tables are
sold?
d. On the same set of axes, graph the manufacturer’s total revenue
and total cost fuctions. Explain how the overhead can be read from the
graph.
Solution
If x is the number of units the manufactured and sold, the total revenue
is given by R( x)  70 x and the total cost by C ( x)  8000  30 x .
a. To find the break-even point, set R ( x ) equal to C ( x) and solve
70x  8000  30x
40x  8000
x  200
so that
It follows that the manufacturer will have to sell 200 units to break
even.
b. The profit P ( x ) is revenue minus cost. Hence
P( x)  R( x)  C ( x)  70 x  (8000  30 x)  40 x  8000
To determine the number of units that must be sold to make a profit of
$6,000, set the formula for profit P ( x ) equal to 6,000 and solve
40x  8000  6000
x  350
so that
It follows that the manufacturer will have to sell 350 units to make a
profit of $6,000.
c. The profit from the sale of 150 tables is
P(150)  40(150)  8000  2000
The minus sign indicates a loss, and it follows that the manufacturer will lose
$2,000 if 150 tables are sold.
R(x)
d.
y
Break-even
C(x)
point
8000
(200, 14000)
x
0
From the graph, we can find that the overhead is the y intercept of C ( x) .
P69
12
In Problem 12, find the indicated limit if it exists.
( x 2  1)(1  2 x) 2
11. xlim
1
Solution
Apply the properties of limits to obtain
lim ( x2  1)(1  2 x)2  lim ( x2  1)  ( lim (1  2 x)) 2  2  32  18
x1
P70
x1
x1
15 , 19 , 23 , 25 , 27 , 31 , 35
In Problems 15, 19, 23 and 25 , find the indicated limit if it exists
15.
x3
lim
x 5 5  x
Solution
The quotient rule for limits does not apply in this case since the limit of
the denominator is
lim(5  x)  0
x 5
x  3)  8 , which is not equal to
Since the limit of the numerator is lim(
x 5
zero, you can conclude that the limit of the quotient does not exist.
19.
lim
x 5
x 2  3x  10
x5
Solution
Both the numerator and the denominator approach 0 as x approach 5.
Simplify the quotient to obtain:
x 2  3x  10 ( x  5)( x  2)

 x2
x 5
x 5
and then take the limit to get
lim
x 5
23.
x 2  3x  10
 lim( x  2)  7
x 5
x 5
x2  x  6
x 2 x 2  3 x  2
lim
Solution
Both the numerator and the denominator approach 0 as x approach -2.
Simplify the quotient to obtain:
x 2  x  6 ( x  2)( x  3) x  3


x 2  3x  2 ( x  2)( x  1) x  1
and then take the limit to get
( x  3)
x2  x  6
x  3 xlim
2

lim

5
x 2 x 2  3 x  2
x 2 x  1
lim ( x  1)
lim
x 2
25.
lim
x4
Solution
x 2
x4
Both the numerator and the denominator approach 0 as x approach 5. to
simplify the quotient, we rationalize the numerator:
x  2 ( x  2)( x  2)
x4



x4
( x  4)( x  2)
( x  4)( x  2)
1
x 2
and then take the limit to get
lim
x 4
x 2
1
1
 lim

x

4
x4
x 2 4
f ( x) and lim f ( x) . If the
For problems 27, 31 and 35, find xlim
 
x  
limiting value is infinite, indicate whether it is   or   .
f ( x) and lim f ( x) . If the
For problems 27, 31 and 35, find xlim
 
x  
limiting value is infinite, indicate whether it is   or   .
27.
f ( x)  x 3  4 x 2  4
Solution
Since the values of this function increase without bound as x increases,
we get
lim ( x3  4 x2  4)  
x
Similarly, we can obtain
lim ( x3  4 x2  4)  
x
31.
f ( x) 
x 2  2x  3
2 x 2  5x  1
Solution
The highest power in the denominator is x 2 . Divide the numerator and
the denominator by x 2 to get
x2  2x  3
1  2 / x  3/ x 2 1  0  0 1

lim


x  2 x 2  5 x  1
x  2  5 / x  1/ x 2
200 2
lim
x2  2x  3
1  2 / x  3/ x 2 1  0  0 1

lim


x  2 x 2  5 x  1
x  2  5 / x  1/ x 2
200 2
lim
35.
f ( x) 
3x 2  6 x  2
2x  9
Solution
The highest power in the denominator is x . Divide the numerator and the
denominator by x to get
3x 2  6 x  2
3x  6  2 / x
 lim
x 
x 
2x  9
29/ x
lim
since
2
lim (3 x  6  )  
x 
x
and
9
lim (2  )  2
x 
x
it follows that
3x 2  6 x  2
 
x 
2x  9
lim
Similarly, we can get
3x 2  6 x  2
3x  6  2 / x
lim
 lim
 
x 
x

2x  9
29/ x
P71
51
51. PER CAPITA EARNINGS Studies indicate that t years form
now, the population of a certiain country will be p  0.2t  1,500 thousand
people, and that gross earnings of the country will be E million dollars,
where
E (t )  9t 2  0.5t  179
a. Express the per capita earnings of the country P  E / p as a
function of time t . (take care with the units.)
b.
What happens to the per capita earnings in the long run (as t   )
Solution
a. Both the gross earnings of the country E and the population p are
functions of time t . Hence the per capita earnings of the country is
9t 2  0.5t  179
P(t )  E (t ) / p(t ) 
0.2t  1500
thousand dollars per person .
b. Since
lim P(t )  lim
t 
t 
9t 2  0.5t  179
9  0.5 / t  179 / t 2
 lim
 15 ,
t 
0.2t  1500
0.2  1500 / t
the per capita earnings tends toward 15 thousand dollar per person
( $15,000 per person) in the long run (as t   ).
P81 11 , 13
In Problems 11 and 13, find the indicated one-sided limit. If the limiting
value is infinite, indicate whether it is
11. lim
x 3

  or   .
x 1  2
x3
Solution
lim
x3
x 1  2
( x  1  2)( x  1  2)
 lim
 lim
x

3
x3
x3
( x  3)( x  1  2)
1
1

x 1  2 4
2 x 2  x
if x  3
3  x
if x  3
13. lim f ( x) and lim f ( x) , where f ( x)  
x 3

x 3

Solution
Since f ( x)  2 x 2  x for x  3 , we have
lim f ( x)  lim (2 x 2  x)  15
x3
Similarly,
x3
f ( x)  3  x for x  3 , so
lim f ( x)  lim (3  x)  0
x3
x 3