MA119-A Applied Calculus for Business
2006 Fall
Practice Problems Review
11/1/2006 & 11/3/2006
1. Packaging By cutting away identical squares from each cornor of a rectangular piece of
cardboard and folding up the resulting ‡aps, the cardboard may be turned into an open
box. If the cardboard is 16 inches long and 10 inches wide, …nd the dimensions of the
box that will yield the maximum volume.
[Solution]
Step 1 Let x denote thelength (in inches) of one side of each of the identical squares to be
cut out of the cardboard and let V denote the volume of the box.
Step 2 The dimension of the box are (16 2x) inches by (10 2x) inches by x inches.
Therefore, the volume is
V = (16
2x) (10
2x) x = 4x3
52x2 + 160x.
Step 3 Since x and each side of the box must be positive, we have x > 0, 16 2x > 0 and
10 2x > 0. This implies that 0 < x < 5.
Step 4 Now, the problem becomes to …nd the maximum volume of V = f (x) = 4x3 52x2 +
160x on the open interval (0; 5).
Step 5 To ensure we have maximum, we consider f (x) is in the close interval [0; 5]. Since
f is continuous on [0; 5], we know that absolute maximum and minimum must be
attained at the endpoints or at the critical points.
Step 6 If f 0 (x) = 12x2 104x + 160 = 0, then we have 4 (3x 20) (x 2) = 0, or, x = 20
3
or 2. But, since 20
> 5, we have the only one critical point x = 2.
3
Step 7 Compare all the values at the endpoints and the critical point. We have f (0) =
f (5) = 0 and f (2) = 144. Thus, the volume of the box is maximized by taking
x = 2. The dimensions of the box are 12" 6" 2" and the volume is 144 cubic
inches.
2. Packaging Betty Moore Company requres that its corned beef hash containers have
a capacity of 54 cubic inches, have the shape of right circular cylinders, and be made
of aluminum. Determine the radius and height of the container that requires the least
amount of metal.
[Solution]
Step 1 Let the radius and height of the container be r and h inches, respectively and let S
denote the surface area of the container.
Step 2 The amount of aluminum used to construct the container is given by the surface area
of the cylinder. The area of the base and the top are each r2 square inches and the
area of the side is 2 rh square inches. Therefore, the surface area is
S = 2 r2 + 2 rh.
1
2
Step 3 The required volume of the container is 54. Thus, we have
r2 h = 54.
Also, we know obviously that r > 0 and h > 0.
Step 4 Now, the problem becomes to …nd the minimize the surface area
S = 2 r2 + 2 rh = 2 r2 + 2 r
54
108
= 2 r2 +
2
r
r
on the open interval (0; 1).
is in the interval (0; 1).
Step 5 We assume that g (r) = 2 r2 + 108
r
108
0
Step 6 If g (r) = 4 r
= 0, then by assuming that r 6= 0, we have 4 r3 108 = 0, or,
r2
3
3
r= p
3 . Thus, we have the only critical point r = p
3 . Note here that r = 0 is not
in our domain since we require that r > 0.
3
3
Step 7 By choosing the test poitns 1 and 3 for the interval 0; p
and p
3
3 ; 1 , respectively, we have g is decreasing from 1 in 0;
Also, we have g 00 (r) = 4 + 216
and g 00
r3
3
test, r = p
is a relative minimum.
3
Step 8 Therefore, we know that when r =
S = 2 r2 +
108
r
3
p
3
=2
2
+
108
3
p
3
3
p
3
3
p
3
3
p
3
and increasing to 1 in
3
p
3
;1 .
37:699 > 0. By the second dericative
2:0484, we have the absolute minimum
79:088.
3 Optimal Subway Fare A city’s Metropolitan Transit Authority (MTA) operates a
subway line for commuters from a certain suburb to the downtown metropolitan area.
Currently, an average of 6000 passengers a day take the trains, paying a fare of $3.00 per
ride. The board of the MTA, contemplating raising the fare to $3.50 per ride in order
to generate a larger revenue, engages the services of a consulting …rm. The …rm’s study
reveals that for each $.50 increase in fare, the ridership will be reduced by an average
of 1000 passengers a day. Thus, the consulting …rm recommends that MTA stick to the
current rate of $3.00 per ride, which already yields a maximum revenue. Show that the
consultants are correct.
[Solution]
Step 1 Let x denote the number of passengers per day, p denote the fare per ride, and R be
MTA’s revenue.
Step 2 Assume that the relationship between x and p is linear (a straight line). The given
data imply that when x = 6000, p = 3, and when x = 5000, p = 3:5. Thus, as points,
(6000; 3) and (5000; 3:5) lie on the line. The slope of the line can be determined by
m=
3:5 3
=
5000 6000
0:0005.
Hence, the equation of the straight line is
(p
3) =
0:0005 (x
6000) ,
or,
p=
0:0005x + 6.
3
Step 3 The revenue per day can also be determined by
R = px =
0:0005x2 + 6x.
Also, we know obviously that p 0 and x 0. By the relationship between p and
x, we can also know that 0:0005x + 6 0, or, x 12000.
Step 4 Now, the problem becomes to …nd the maximize the revenue
R=
0:0005x2 + 6x
on the closed interval [0; 12000].
Step 5 We assume that f (r) = 0:0005x2 + 6x is in the interval [0; 12000]. Since f is
continuous on [0; 12000], we know that absolute maximum and minimum must be
attained at the endpoints or at the critical points.
Step 6 If f 0 (x) = 0:001x + 6 = 0, then we have x = 6000. Thus, we have the only critical
point x = 6000.
Step 7 Compare all the values at the endpoints and the critical point. We have f (0) =
f (12000) = 0 and f (6000) = 18000. Thus, the revenue is maximized when x = 6000.
Step 8 Therefore, the maximum revenue $18000 per day is realized when the ridership is
6000 per day. The optimum price of the fare per ride is $3:00.
4. Inventory Control and Planning Dixie Import-Export is the sole agent for the Excalibur 250-cc motorcycle. Management estimates that the demand for these motorcycles
is 10,000 per year and that they will sell at a uniform rate throught the year. The cost
incurred in ordering each shipment of motorcycles is $10,000, and the cost per year of
storng each motorcycle is $200.
Dixie’s management faces the following problem: Oedering too many motorcycles at
one time tie up valuable storage space and increases the storage cost. On the other hand,
placing orders too frequently increases the ordering costs. How large should each order
be, and how often should orders be placed, to minimize ordering and storage costs?
[Solution]
Step 1 Let x denote the number of motorcycles in each order.
Step 2 Assume that each shipment arrives just as the previous shipment has been sold, the
average number of motorcycles in storage during the year is x2 . The cost of storage
x
for the year is 200
= 100x dollars for the year.
2
Step 3 Since the company requires 10,000 motorcycles for the year and since each order is
. Thus, the ordering cost is
for x mototcycles, the number of orders required is 10;000
x
10;000
100;000;000
=
10; 000
dollars for the year.
x
x
Step 4 The total cost is
100; 000; 000
C (x) = 100x +
.
x
Step 5 Obviously, x > 0 and x 10000. Now, the problem becomes to …nd the minimize
the cost
100; 000; 000
C (x) = 100x +
x
on the closed interval (0; 10000].
Step 6 If C 0 (x) = 100 100;000;000
= 0, then we have x2 = 1; 000; 000, or x = 1000. Since
x2
x = 1000 is not in the domain, we have the only critical point x = 1000.
4
Step 7 By choosing the test poitns 1 and 5000 for the interval (0; 1000) and (1000; 10000],
respectively, we have f is decreasing from 1 in (0; 1000) and increasing to 1 in
(1000; 10000]. Also, we have C 00 (r) = 200;000;000
and C 00 (1000) = 0:2 > 0. By the
x3
second dericative test, x = 1000 is a relative minimum.
Step 8 Therefore, the size of order x = 1000 should be placed to minimize the cost.
5. Monthly Mortgage Payment You decide to purchase a 3 bedroom townhouse in Easton at thr price $352,000. The mortgage agent o¤er you a 30 years …xed rate mortgage at
the annual interest rate 6% if you can pay 20% down payment due at signing. The interest
is compounded monthly in the payment day. What will be your monthly payment?
[First Thought]
The down payment is 352000 0:2 = 70400. Thus, the loan value is 352000 70400 =
281600. With a 30 years …xed rate mortgage at the annual interest rate 6% compounded
monthly, the total amount after 30 years is
A=P 1+
r
m
mt
= 281600
1+
0:06
12
12 30
= 1696000.
In 30 years, there will be 30 12 = 360 payment day of we make out payment monthly.
So, the monthly payment is 1696000
= 4711:10.
360
[Solution]
Assume that the loan value is L, the annual rate is r and you pay m times a year for
t years. Thus, there are mt payments to pay. Assume that the monthly payment is P .
We use Li where i = 1; 2;
; mt to represent the load value right after the ith payment.
Thus, L0 = L. Also, Ii denotes the interest compounded in the ith payment day.
In the 1st payment day, the interest I1 = L0 mr . Since the payment goes to the
interest …rst, we pay only P I1 to the loan value. Thus, the load value becomes
L1 = L0
I1 = L0
(P
I1 ) = L0 + L0
r
m
P = L0 1 +
r
m
P.
In the 2nd payment day, the interest I2 = L1 mr . Since the payment goes to the interest
…rst, we pay only P I2 to the loan value. Thus, the load value becomes
L2 = L1
I2 = L1
(P
Similarly, we have Li = Li
We observe that
r
m
r
1+
m
r
1+
m
I2 ) = L1 + L1
1
L2 = L1 1 +
= L0
= L0
1+
r
m
P
r
m
r
P
m
r
1+
+1
m
P 1+
2
P = L1 1 +
P for all i = 1; 2;
P = L0 1 +
2
r
m
P
1+
r
m
P.
; mt.
r
m
P
5
and
r
m
r
1+
m
r
1+
m
L 3 = L2 1 +
= L0
= L0
P =
L0 1 +
2
P
1+
r
+1
m
1+
r
m
P
r
P
m
r 2
r
1+
+ 1+
+1 .
m
m
3
P 1+
3
P
r
m
r
m
2
P 1+
By mathematical induction, we have
Li = L0 1 +
Also, since 1 +
r
1+
m
i 1
r
m
r
m
i
P
1+
r
m
i 1
+ 1+
r
m
i 2
+
+ 1+
r
+1 .
m
> 1, we have
r
+ 1+
m
i 2
+
1 + mr
r
+1=
+ 1+
m
1 + mr
Thus,
Li = L0
r
1+
m
1+
i
P
r i
m
r
m
1
i
!
1
1+
=
1
r i
m
r
m
1
.
.
Our monthly payment should pay o¤ the loan by mt payments. Thus, we can solve for
P by setting Lmt = 0.
In our case, the down payment is 352000 0:2 = 70400. Thus, the loan value is
352000 70400 = 281600. In 30 years, there will be 30 12 = 360 payment days if we
make our payment monthly. With a 30 years …xed rate mortgage at the annual interest
rate 6% compounded monthly, the total amount after 30 years is
!
360
360
1 + 0:06
1
0:06
12
0 = L360 = 281600 1 +
P
0:06
12
12
= 1696000
1004:5P .
So, the monthly payment is P =
1696000
1004:5
= 1688:4.
6. Real Estate Investments A condominium complex was purchases by a group of private
investors for $1.4 million and sold 6 yr later for $3.6 million. Find the annual rate of
return (compounded continuously) on their investment.
[Solution]
Let the annual rate of return is r. We compound continuously. Thus, we have
3:6 = 1:4ert .
Therefore,
3:6
.
1:4
By applying ln on the both sides, we have
e6r =
6r = ln e6r = ln
3:6
1:4
.
6
Hence,
r=
1
ln
6
3:6
1:4
= 0:15741.
So, the annual rate of return is 15:741%.
7. Growth of Bacteria Under ideal laboratory conditions, the number of bacteria in a
culture grows in accordance with the law Q (t) = Q0 ekt , where Q0 denotes the number
of bacteria initially present in the culture, k is a constant determined by the strain of
bacteria under consideration, and t is the elapsed time measured in hours. Suppose 10,000
bacteria are present initially in the culture and 60,000 present 2 hours later.
(a) How many bacteria will there be in the culture at the end of 4 hours?
(b) What is the rate of growth of the population after 4 hours?
[Solution]
(a) Since 10,000 bacteria are present initially in the culture, we have Q0 = 10000. Also,
2 hours later, we have 60000 bacteria. This tells us that 60000 = Q (2) = Q0 e2k =
10000e2k , or, 6 = e2k .
To solve k, we apply ln on the both sides to get
ln 6 = ln e2k = 2k ln e = 2k.
Thus, k = ln26 .
At the end of 4 hours, there will be
Q (4) = 10000e4
ln 6
2
2
= 10000e2 ln 6 = 10000eln 6 = 10000
62 = 360000.
bacteria.
(b) The rate of growth is Q0 (t) = Q0 kekt . After 4 hours,
Q0 (4) = 10000
ln 6
2
e4
ln 6
2
= 5000
e2 ln 6 = 180000 ln 6
ln 6
322520.
8. Carbon-14 Dating A skull from an archeological site has one-tenth the amount of C-14
that it originally contained. Determine the approximate age of the skull.
[Solution]
The Carbon 14 decays exponentially by the formula Q (t) = Q0 e kt where k is the
decay constant. Since the half-life of C-14 is 5770 years, we have
Q0
= Q (5770) = Q0 e
2
ln
5770k
.
1
ln 2
2
This tells us that e 5770k = 12 , or, k = 5770
= 5770
0:00012013.
Since there are only one-tenth the amount of C-14 left in the skull, we have
Q0
= Q (t) = Q0 e
10
ln 2
ln
1
1
This tells us that e 5770 t = 10
, or, k = ln102 =
5770
of the skull is approximately 19168 years.
ln 2
t
5770
.
ln 10
ln 2
5770
=
5770 ln 10
ln 2
= 19168. Thus, the age
7
9. Assembly Time The Camera Division of Eastman Optical produces a 35-mm single lens
re‡ex camera. Eastman’s training department determines that after completing the basic
training program, a new, previously inexperienced employee will be able to assemble
Q (t) = 50
30e
0:5t
.
model F cameras per day, t months after the employee starts work on the assembly line.
(a) How many model F cameras can a new employee assemble per day after basic training?
(b) How many model F cameras can an employee with 1 month of experience assemble
per day? An exployee with 2 months experience? An exployee with 6 months
experience?
(c) How many model F cameras can the average experienced employee assemble per
day?
[Solution]
(a) The number of model F cameras a new employee can assemble per day is given by
Q (0) = 50 30 = 20.
(b) The number of model F cameras that an employee with 1 month of experience can
assemble per day is given by Q (1) = 50 30e 0:5(1) 31:804.
The number of model F cameras that an employee with 2 months of experience can
assemble per day is given by Q (2) = 50 30e 0:5(2) 38; 964.
The number of model F cameras that an employee with 6 months of experience can
assemble per day is given by Q (6) = 50 30e 0:5(6) 48:506.
(c) As t ! 1, by the graph of the natural exponential function,
lim e
t!1
0:5t
= 0.
Therefore, the average experienced employee can assemble per day is 50.
10. Spread of Flu The number of soldiers at Fort MacArthur who contracted in‡uenza after
t days during a ‡u epidemic is approximated by the exponential model
Q (t) =
5000
1 + 1249e
kt
.
If 40 soldiers contracted the ‡u by day 7, …nd how many soliders contracted the ‡u by
day 15.
[Solution]
Since there are 40 soldiers contracted the ‡u by day 7, we have
40 = Q (7) =
5000
1 + 1249e
k(7)
.
Thus,
40 1 + 1249e 7k
1 + 1249e 7k
1249e 7k
e 7k
7k = ln e 7k
k
= 5000
= 5000
= 125
40
= 124
124
= 1249
124
= ln 1249
= ln 124 ln 1249
ln 124 ln 1249
=
0:32997
7
8
Therefore, the number of soldiers who contracted in‡uenza after t days is
5000
Q (t) =
.
ln 124 ln 1249
)t
7
1 + 1249e(
By day 15, the number of soldiers who contracted in‡uenza is
5000
Q (15) =
507:58,
ln 124 ln 1249
)15
7
1 + 1249e(
or, approximately 508 soldiers.
11. In‡ation Rate of an Economy The function
I (t) =
0:2t3 + 3t2 + 100
(0
t
9)
gives the CPI of an economy, where t = 0 corresponds to the beginning of 1995.
(a) Find the in‡ation rate at the beginning of 2001 (t = 6).
(b) Show that in‡ation was moderating at that time.
[Solution]
Suppose the consumer price index (CPI) of an economy between the years a and b is
described the function I (t) (a
t
b). Then the …rst derivative of I at t = c, I 0 (c),
where a < c < b, gives the rate of change of I at c. The quantity
I 0 (c)
I (c)
is called the relative rate of change of I (t) with respect to t at t = c, measures the
in‡ation rate of the economy at t = c. The second derivative of I at t = c, I 00 (c) gives
the rate of change of I 0 at t = c.
If, at t = c, I 0 (c) is positive and I 00 (c) is negative, then we say that the economy is
experiencing in‡ation (the CPI is increasing) but the rate at which in‡ation is growing is
in fact decreasing. This is the situation describled by an economist "in‡ation is slowing."
(a) We …nd I 0 (t) = 0:6t2 + 6t. Also, we compute that I 0 (6) = 14:4 and I (6) = 164:8.
Thus, the in‡ation rate at the beginning of 2001 is
I 0 (6)
14:4
=
0:087379,
I (6)
164:8
or, 8:7%.
(b) We have I 00 (t) = 1:2t + 6. Thus, I 00 (6) = 1:2. Therefore, we can see that I 0 is
indeed decreasing at t = 6. Hence, we conclude that the in‡ation was moderating at
that time.
12. Watching a Rocket Launch At a distance of 4000 feet from the launch site, a spectator
is observing a rocket being launched. If the rocket lifts o¤ vertically and is rising at a
speed of 600 feet/second when it is at an altitude of 3000 feet, how fast is the distance
between the rocket and the spectator changing at that instant?
[Solution]
Step 1 Let y = y (t) be the altitude of the rocket and x = x (t) be the distance between the
rocket and the spectator at any time t.
Step 2 When the rocket is in the 3000 feet high, y = 3000, its speed is 600 feet/second,
dy
= 600.
dt
9
Step 3 The problem is looking for how fast is the distance between the rocket anf the
spectator changing at that instant. Thus, dx
.
dt
Step 4 By the Pythagorean theorem, we have
p
x2 = y 2 + 40002 .
Thus, we have x = 30002 + 40002 = 5000.
Step 5 Now, by di¤erentiating the equation on both sides with respect to t, we get
dy
dx
= 2y .
2x
dt
dt
Step 6 By subsititing x = 5000, y = 3000, and dy
= 600, we can solve for
dt
dx
2 3000 600
=
= 360.
dt
2 5000
Step 7 Therefore, the distance between the rocket and the spectator is changing at the rate
360 feet/second.