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Section 4.3: Rational Inequalities and Applications, from College Algebra: Corrected Edition by Carl Stitz,
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342 Rational Functions
4.3 Rational Inequalities and Applications
In this section, we solve equations and inequalities involving rational functions and explore associ-
ated application problems. Our first example showcases the critical difference in procedure between
solving a rational equation and a rational inequality.
Example 4.3.1.
1. Solve
x3 2x + 1
x 1
=
1
2
x 1. 2. Solve
x3 2x + 1
x 1
1
2
x 1.
3. Use your calculator to graphically check your answers to 1 and 2.
Solution.
1. To solve the equation, we clear denominators
x3 2x + 1
x 1
=
1
2
x 1(
x3 2x + 1
x 1
)
2(x 1) =
(
1
2
x 1
)
2(x 1)
2×3 4x + 2 = x2 3x + 2 expand
2×3 x2 x = 0
x(2x + 1)(x 1) = 0 factor
x = 1
2
, 0, 1
Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we
see that x = 1 does not satisfy the original equation and must be discarded. Our solutions
are x = 1
2
and x = 0.
2. To solve the inequality, it may be tempting to begin as we did with the equation namely
by multiplying both sides by the quantity (x 1). The problem is that, depending on x,
(x 1) may be positive (which doesnt affect the inequality) or (x 1) could be negative
(which would reverse the inequality). Instead of working by cases, we collect all of the terms
on one side of the inequality with 0 on the other and make a sign diagram using the technique
given on page 321 in Section 4.2.
x3 2x + 1
x 1
1
2
x 1
x3 2x + 1
x 1
1
2
x + 1 0
2
(
x3 2x + 1
)
x(x 1) + 1(2(x 1))
2(x 1)
0 get a common denominator
2×3 x2 x
2x 2
0 expand
4.3 Rational Inequalities and Applications 343
Viewing the left hand side as a rational function r(x) we make a sign diagram. The only
value excluded from the domain of r is x = 1 which is the solution to 2x 2 = 0. The zeros
of r are the solutions to 2×3 x2 x = 0, which we have already found to be x = 0, x = 1
2
and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test
values in each test interval, we construct the sign diagram below.
1
2
0 1
(+) 0 () 0 (+) (+)
We are interested in where r(x) 0. We find r(x) > 0, or (+), on the intervals
(
,1
2
)
,
(0, 1) and (1,). We add to these intervals the zeros of r, 1
2
and 0, to get our final solution:(
,1
2
]
[0, 1) (1,).
3. Geometrically, if we set f(x) = x
32x+1
x1 and g(x) =
1
2
x 1, the solutions to f(x) = g(x) are
the x-coordinates of the points where the graphs of y = f(x) and y = g(x) intersect. The
solution to f(x) g(x) represents not only where the graphs meet, but the intervals over
which the graph of y = f(x) is above (>) the graph of g(x). We obtain the graphs below.
The Intersect command confirms that the graphs cross when x = 1
2
and x = 0. It is clear
from the calculator that the graph of y = f(x) is above the graph of y = g(x) on
(
,1
2
)
as well as on (0,). According to the calculator, our solution is then
(
,1
2
]
[0,)
which almost matches the answer we found analytically. We have to remember that f is not
defined at x = 1, and, even though it isnt shown on the calculator, there is a hole1 in the
graph of y = f(x) when x = 1 which is why x = 1 is not part of our final answer.
Next, we explore how rational equations can be used to solve some classic problems involving rates.
Example 4.3.2. Carl decides to explore the Meander River, the location of several recent Sasquatch
sightings. From camp, he canoes downstream five miles to check out a purported Sasquatch nest.
Finding nothing, he immediately turns around, retraces his route (this time traveling upstream),
1There is no asymptote at x = 1 since the graph is well behaved near x = 1. According to Theorem 4.1, there
must be a hole there.
344 Rational Functions
and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water,
how fast was the Meander River flowing on that day?
Solution. We are given information about distances, rates (speeds) and times. The basic principle
relating these quantities is:
distance = rate time
The first observation to make, however, is that the distance, rate and time given to us arent
compatible: the distance given is the distance for only part of the trip, the rate given is the speed
Carl can canoe in still water, not in a flowing river, and the time given is the duration of the entire
trip. Ultimately, we are after the speed of the river, so lets call that R measured in miles per hour
to be consistent with the other rate given to us. To get started, lets divide the trip into its two
parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we
know is that the distance traveled is 5 miles.
distance downstream = rate traveling downstream time traveling downstream
5 miles = rate traveling downstream time traveling downstream
Since the return trip upstream followed the same route as the trip downstream, we know that the
distance traveled upstream is also 5 miles.
distance upstream = rate traveling upstream time traveling upstream
5 miles = rate traveling upstream time traveling upstream
We are told Carl can canoe at a rate of 6 miles per hour in still water. How does this figure
into the rates traveling upstream and downstream? The speed the canoe travels in the river is a
combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and
the speed of the river, which were calling R. When traveling downstream, the river is helping Carl
along, so we add these two speeds:
rate traveling downstream = rate Carl propels the canoe + speed of the river
= 6 miles
hour
+ Rmiles
hour
So our downstream speed is (6 + R) miles
hour
. Substituting this into our distance-rate-time equation
for the downstream part of the trip, we get:
5 miles = rate traveling downstream time traveling downstream
5 miles = (6 + R) miles
hour
time traveling downstream
When traveling upstream, Carl works against the current. Since the canoe manages to travel
upstream, the speed Carl can canoe in still water is greater than the rivers speed, so we subtract
the rivers speed from Carls canoing speed to get:
rate traveling upstream = rate Carl propels the canoe river speed
= 6 miles
hour
Rmiles
hour
Proceeding as before, we get
4.3 Rational Inequalities and Applications 345
5 miles = rate traveling upstream time traveling upstream
5 miles = (6 R) miles
hour
time traveling upstream
The last piece of information given to us is that the total trip lasted 3 hours. If we let tdown denote
the time of the downstream trip and tup the time of the upstream trip, we have: tdown+tup = 3 hours.
Substituting tdown and tup into the distance-rate-time equations, we get (suppressing the units)
three equations in three unknowns:2
E1 (6 + R) tdown = 5
E2 (6 R) tup = 5
E3 tdown + tup = 3
Since we are ultimately after R, we need to use these three equations to get at least one equation
involving only R. To that end, we solve E1 for tdown by dividing both sides
3 by the quantity (6 +R)
to get tdown =
5
6+R
. Similarly, we solve E2 for tup and get tup =
5
6R. Substituting these into E3,
we get:4
5
6 + R
+
5
6 R
= 3.
Clearing denominators, we get 5(6 R) + 5(6 + R) = 3(6 + R)(6 R) which reduces to R2 = 16.
We find R = 4, and since R represents the speed of the river, we choose R = 4. On the day in
question, the Meander River is flowing at a rate of 4 miles per hour.
One of the important lessons to learn from Example 4.3.2 is that speeds, and more generally, rates,
are additive. As we see in our next example, the concept of rate and its associated principles can
be applied to a wide variety of problems – not just distance-rate-time scenarios.
Example 4.3.3. Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can
weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own?
Solution. The key relationship between work and time which we use in this problem is:
amount of work done = rate of work time spent working
We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylors case then:
amount of work Taylor does = rate of Taylor working time Taylor spent working
1 garden = (rate of Taylor working) (4 hours)
So we have that the rate Taylor works is
1 garden
4 hours
= 1
4
garden
hour
. We are also told that when working
together, Taylor and Carl can weed the garden in just 3 hours. We have:
2This is called a system of equations. No doubt, youve had experience with these things before, and we will study
systems in greater detail in Chapter 8.
3While we usually discourage dividing both sides of an equation by a variable expression, we know (6 + R) 6= 0
since otherwise we couldnt possibly multiply it by tdown and get 5.
4The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started,
the units on the 3 is hours.
346 Rational Functions
amount of work done together = rate of working together time spent working together
1 garden = (rate of working together) (3 hours)
From this, we find that the rate of Taylor and Carl working together is
1 garden
3 hours
= 1
3
garden
hour
. We are
asked to find out how long it would take for Carl to weed the garden on his own. Let us call this
unknown t, measured in hours to be consistent with the other times given to us in the problem.
Then:
amount of work Carl does = rate of Carl working time Carl spent working
1 garden = (rate of Carl working) (t hours)
In order to find t, we need to find the rate of Carl working, so lets call this quantity R, with units
garden
hour
. Using the fact that rates are additive, we have:
rate working together = rate of Taylor working + rate of Carl working
1
3
garden
hour
= 1
4
garden
hour
+ R
garden
hour
so that R = 1
12
garden
hour
. Substituting this into our work-rate-time equation for Carl, we get:
1 garden = (rate of Carl working) (t hours)
1 garden =
(
1
12
garden
hour
)
(t hours)
Solving 1 = 1
12
t, we get t = 12, so it takes Carl 12 hours to weed the garden on his own.5
As is common with word problems like Examples 4.3.2 and 4.3.3, there is no short-cut to the
answer. We encourage the reader to carefully think through and apply the basic principles of rate
to each (potentially different!) situation. It is time well spent. We also encourage the tracking of
units, especially in the early stages of the problem. Not only does this promote uniformity in the
units, it also serves as a quick means to check if an equation makes sense.6
Our next example deals with the average cost function, first introduced on page 82, as applied to
PortaBoy Game systems from Example 2.1.5 in Section 2.1.
Example 4.3.4. Given a cost function C(x), which returns the total cost of producing x items,
recall that the average cost function, C(x) =
C(x)
x
computes the cost per item when x items are
produced. Suppose the cost C, in dollars, to produce x PortaBoy game systems for a local retailer
is C(x) = 80x + 150, x 0.
1. Find an expression for the average cost function C(x).
2. Solve C(x) < 100 and interpret. 5Carl would much rather spend his time writing open-source Mathematics texts than gardening anyway. 6In other words, make sure you dont try to add apples to oranges! 4.3 Rational Inequalities and Applications 347 3. Determine the behavior of C(x) as x and interpret. Solution. 1. From C(x) = C(x) x , we obtain C(x) = 80x+150 x . The domain of C is x 0, but since x = 0 causes problems for C(x), we get our domain to be x > 0, or (0,).
2. Solving C(x) < 100 means we solve 80x+150
x
< 100. We proceed as in the previous example.
80x + 150
x
< 100
80x + 150
x
100 < 0
80x + 150 100x
x
< 0 common denominator
150 20x
x
< 0
If we take the left hand side to be a rational function r(x), we need to keep in mind that the
applied domain of the problem is x > 0. This means we consider only the positive half of the
number line for our sign diagram. On (0,), r is defined everywhere so we need only look
for zeros of r. Setting r(x) = 0 gives 150 20x = 0, so that x = 15
2
= 7.5. The test intervals
on our domain are (0, 7.5) and (7.5,). We find r(x) < 0 on (7.5,).
0
7.5
(+) 0 ()
In the context of the problem, x represents the number of PortaBoy games systems produced
and C(x) is the average cost to produce each system. Solving C(x) < 100 means we are trying
to find how many systems we need to produce so that the average cost is less than $100 per
system. Our solution, (7.5,) tells us that we need to produce more than 7.5 systems to
achieve this. Since it doesnt make sense to produce half a system, our final answer is [8,).
3. When we apply Theorem 4.2 to C(x) we find that y = 80 is a horizontal asymptote to
the graph of y = C(x). To more precisely determine the behavior of C(x) as x , we
first use long division7 and rewrite C(x) = 80 + 150
x
. As x , 150
x
0+, which means
C(x) 80 + very small (+). Thus the average cost per system is getting closer to $80 per
system. If we set C(x) = 80, we get 150
x
= 0, which is impossible, so we conclude that
C(x) > 80 for all x > 0. This means that the average cost per system is always greater than
$80 per system, but the average cost is approaching this amount as more and more systems
are produced. Looking back at Example 2.1.5, we realize $80 is the variable cost per system
7In this case, long division amounts to term-by-term division.
348 Rational Functions
the cost per system above and beyond the fixed initial cost of $150. Another way to interpret
our answer is that infinitely many systems would need to be produced to effectively zero
out the fixed cost.
Our next example is another classic box with no top problem.
Example 4.3.5. A box with a square base and no top is to be constructed so that it has a volume
of 1000 cubic centimeters. Let x denote the width of the box, in centimeters as seen below.
width, x
height
depth
1. Express the height h in centimeters as a function of the width x and state the applied domain.
2. Solve h(x) x and interpret.
3. Find and interpret the behavior of h(x) as x 0+ and as x .
4. Express the surface area S of the box as a function of x and state the applied domain.
5. Use a calculator to approximate (to two decimal places) the dimensions of the box which
minimize the surface area.
Solution.
1. We are told that the volume of the box is 1000 cubic centimeters and that x represents the
width, in centimeters. From geometry, we know Volume = widthheightdepth. Since the
base of the box is a square, the width and the depth are both x centimeters. Using h for the
height, we have 1000 = x2h, so that h = 1000
x2
. Using function notation,8 h(x) = 1000
x2
As for
the applied domain, in order for there to be a box at all, x > 0, and since every such choice
of x will return a positive number for the height h we have no other restrictions and conclude
our domain is (0,).
2. To solve h(x) x, we proceed as before and collect all nonzero terms on one side of the
inequality in order to use a sign diagram.
8That is, h(x) means h of x, not h times x here.
4.3 Rational Inequalities and Applications 349
h(x) x
1000
x2
x
1000
x2
x 0
1000 x3
x2
0 common denominator
We consider the left hand side of the inequality as our rational function r(x). We see r is
undefined at x = 0, but, as in the previous example, the applied domain of the problem is
x > 0, so we are considering only the behavior of r on (0,). The sole zero of r comes when
1000×3 = 0, which is x = 10. Choosing test values in the intervals (0, 10) and (10,) gives
the following diagram.
0
(+)
10
0 ()
We see r(x) > 0 on (0, 10), and since r(x) = 0 at x = 10, our solution is (0, 10]. In the context
of the problem, h represents the height of the box while x represents the width (and depth)
of the box. Solving h(x) x is tantamount to finding the values of x which result in a box
where the height is at least as big as the width (and, in this case, depth.) Our answer tells
us the width of the box can be at most 10 centimeters for this to happen.
3. As x 0+, h(x) = 1000
x2
. This means that the smaller the width x (and, in this
case, depth), the larger the height h has to be in order to maintain a volume of 1000 cubic
centimeters. As x , we find h(x) 0+, which means that in order to maintain a volume
of 1000 cubic centimeters, the width and depth must get bigger as the height becomes smaller.
4. Since the box has no top, the surface area can be found by adding the area of each of the
sides to the area of the base. The base is a square of dimensions x by x, and each side has
dimensions x by h. We get the surface area, S = x2 + 4xh. To get S as a function of x, we
substitute h = 1000
x2
to obtain S = x2 + 4x
(
1000
x2
)
. Hence, as a function of x, S(x) = x2 + 4000
x
.
The domain of S is the same as h, namely (0,), for the same reasons as above.
5. A first attempt at the graph of y = S(x) on the calculator may lead to frustration. Chances
are good that the first window chosen to view the graph will suggest y = S(x) has the x-axis
as a horizontal asymptote. From the formula S(x) = x2 + 4000
x
, however, we get S(x) x2 as
x , so S(x) . Readjusting the window, we find S does possess a relative minimum
at x 12.60. As far as we can tell,9 this is the only relative extremum, so it is the absolute
minimum as well. This means that the width and depth of the box should each measure
9without Calculus, that is…
350 Rational Functions
approximately 12.60 centimeters. To determine the height, we find h(12.60) 6.30, so the
height of the box should be approximately 6.30 centimeters.
4.3.1 Variation
In many instances in the sciences, rational functions are encountered as a result of fundamental
natural laws which are typically a result of assuming certain basic relationships between variables.
These basic relationships are summarized in the definition below.
Definition 4.5. Suppose x, y and z are variable quantities. We say
y varies directly with (or is directly proportional to) x if there is a constant k such
that y = kx.
y varies inversely with (or is inversely proportional to) x if there is a constant k
such that y = k
x
.
z varies jointly with (or is jointly proportional to) x and y if there is a constant k
such that z = kxy.
The constant k in the above definitions is called the constant of proportionality.
Example 4.3.6. Translate the following into mathematical equations using Definition 4.5.
1. Hookes Law: The force F exerted on a spring is directly proportional the extension x of the
spring.
2. Boyles Law: At a constant temperature, the pressure P of an ideal gas is inversely propor-
tional to its volume V .
3. The volume V of a right circular cone varies jointly with the height h of the cone and the
square of the radius r of the base.
4. Ohms Law: The current I through a conductor between two points is directly proportional to
the voltage V between the two points and inversely proportional to the resistance R between
the two points.
4.3 Rational Inequalities and Applications 351
5. Newtons Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass
M, are positioned so that the distance between their centers of mass is r. The gravitational
force F exerted on the two objects varies directly with the product of the two masses and
inversely with the square of the distance between their centers of mass.
Solution.
1. Applying the definition of direct variation, we get F = kx for some constant k.
2. Since P and V are inversely proportional, we write P = k
V
.
3. There is a bit of ambiguity here. Its clear that the volume and the height of the cone are
represented by the quantities V and h, respectively, but does r represent the radius of the base
or the square of the radius of the base? It is the former. Usually, if an algebraic operation is
specified (like squaring), it is meant to be expressed in the formula. We apply Definition 4.5
to get V = khr2.
4. Even though the problem doesnt use the phrase varies jointly, it is implied by the fact that
the current I is related to two different quantities. Since I varies directly with V but inversely
with R, we write I = kV
R
.
5. We write the product of the masses mM and the square of the distance as r2. We have that
F varies directly with mM and inversely with r2, so F = kmM
r2
.
In many of the formulas in the previous example, more than two varying quantities are related. In
practice, however, usually all but two quantities are held constant in an experiment and the data
collected is used to relate just two of the variables. Comparing just two varying quantities allows
us to view the relationship between them as functional, as the next example illustrates.
Example 4.3.7. According to this website the actual data relating the volume V of a gas and its
pressure P used by Boyle and his assistant in 1662 to verify the gas law that bears his name is
given below.
V 48 46 44 42 40 38 36 34 32 30 28 26 24
P 29.13 30.56 31.94 33.5 35.31 37 39.31 41.63 44.19 47.06 50.31 54.31 58.81
V 23 22 21 20 19 18 17 16 15 14 13 12
P 61.31 64.06 67.06 70.69 74.13 77.88 82.75 87.88 93.06 100.44 107.81 117.56
1. Use your calculator to generate a scatter diagram for these data using V as the independent
variable and P as the dependent variable. Does it appear from the graph that P is inversely
proportional to V ? Explain.
2. Assuming that P and V do vary inversely, use the data to approximate the constant of
proportionality.
352 Rational Functions
3. Use your calculator to determine a Power Regression for this data10 and use it verify your
results in 1 and 2.
Solution.
1. If P really does vary inversely with V , then P = k
V
for some constant k. From the data plot,
the points do seem to lie along a curve like y = k
x
.
2. To determine the constant of proportionality, we note that from P = k
V
, we get k = PV .
Multiplying each of the volume numbers times each of the pressure numbers,11 we produce a
number which is always approximately 1400. We suspect that P = 1400
V
. Graphing y = 1400
x
along with the data gives us good reason to believe our hypotheses that P and V are, in fact,
inversely related.
The graph of the data The data with y = 1400
x
3. After performing a Power Regression, the calculator fits the data to the curve y = axb where
a 1400 and b 1 with a correlation coefficient which is darned near perfect.12 In other
words, y = 1400×1 or y = 1400
x
, as we guessed.
10We will talk more about this in the coming chapters.
11You can use tell the calculator to do this arithmetic on the lists and save yourself some time.
12We will revisit this example once we have developed logarithms in Chapter 6 to see how we can actually linearize
this data and do a linear regression to obtain the same result.
4.3 Rational Inequalities and Applications 353
4.3.2 Exercises
In Exercises 1 – 6, solve the rational equation. Be sure to check for extraneous solutions.
1.
x
5x + 4
= 3 2.
3x 1
x2 + 1
= 1
3.
1
x + 3
+
1
x 3
=
x2 3
x2 9
4.
2x + 17
x + 1
= x + 5
5.
x2 2x + 1
x3 + x2 2x
= 1 6.
x3 + 4x
x2 9
= 4x
In Exercises 7 – 20, solve the rational inequality. Express your answer using interval notation.
7.
1
x + 2
0 8.
x 3
x + 2
0 9.
x
x2 1
> 0
10.
4x
x2 + 4
0 11.
x2 x 12
x2 + x 6
> 0 12.
3×2 5x 2
x2 9
< 0
13.
x3 + 2x2 + x
x2 x 2
0 14.
x2 + 5x + 6
x2 1
> 0 15.
3x 1
x2 + 1
1
16.
2x + 17
x + 1
> x + 5 17.
x3 + 4x
x2 9
4x 18.
1
x2 + 1
< 0
19.
x4 4x3 + x2 2x 15
x3 4x2
x 20.
5x3 12x2 + 9x + 10
x2 1
3x 1
21. Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour
and finishes the race 10 minutes before Carl, how fast does Carl run?
22. One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow
nesting near Donnies house flies three quarters of a mile down the road (in the direction of
the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the
unladen swallow? (Here, airspeed is the speed that the swallow can fly in still air.)
23. In order to remove water from a flooded basement, two pumps, each rated at 40 gallons per
minute, are used. After half an hour, the one pump burns out, and the second pump finishes
removing the water half an hour later. How many gallons of water were removed from the
basement?
24. A faucet can fill a sink in 5 minutes while a drain will empty the same sink in 8 minutes. If
the faucet is turned on and the drain is left open, how long will it take to fill the sink?
25. Working together, Daniel and Donnie can clean the llama pen in 45 minutes. On his own,
Daniel can clean the pen in an hour. How long does it take Donnie to clean the llama pen on
his own?
354 Rational Functions
26. In Exercise 32, the function C(x) = .03x3 4.5x2 + 225x + 250, for x 0 was used to model
the cost (in dollars) to produce x PortaBoy game systems. Using this cost function, find the
number of PortaBoys which should be produced to minimize the average cost C. Round your
answer to the nearest number of systems.
27. Suppose we are in the same situation as Example 4.3.5. If the volume of the box is to be 500
cubic centimeters, use your calculator to find the dimensions of the box which minimize the
surface area. What is the minimum surface area? Round your answers to two decimal places.
28. The box for the new Sasquatch-themed cereal, Crypt-Os, is to have a volume of 140 cubic
inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the
base of the box.13 Find the dimensions of the box which will minimize the surface area of the
box. What is the minimum surface area? Round your answers to two decimal places.
29. Sally is Skippys neighbor from Exercise 19 in Section 2.3. Sally also wants to plant a vegetable
garden along the side of her home. She doesnt have any fencing, but wants to keep the size
of the garden to 100 square feet. What are the dimensions of the garden which will minimize
the amount of fencing she needs to buy? What is the minimum amount of fencing she needs
to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will
border the house, Sally doesnt need fencing along that side.)
30. Another Classic Problem: A can is made in the shape of a right circular cylinder and is to
hold one pint. (For dry goods, one pint is equal to 33.6 cubic inches.)14
(a) Find an expression for the volume V of the can in terms of the height h and the base
radius r.
(b) Find an expression for the surface area S of the can in terms of the height h and the
base radius r. (Hint: The top and bottom of the can are circles of radius r and the side
of the can is really just a rectangle that has been bent into a cylinder.)
(c) Using the fact that V = 33.6, write S as a function of r and state its applied domain.
(d) Use your graphing calculator to find the dimensions of the can which has minimal surface
area.
31. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius
of the base and height) of the drum which would minimize the surface area. What is the
minimum surface area? Round your answers to two decimal places.
32. In Exercise 71 in Section 1.4, the population of Sasquatch in Portage County was modeled
by the function P(t) = 150t
t+15
, where t = 0 represents the year 1803. When were there fewer
than 100 Sasquatch in Portage County?
131.62 is a crude approximation of the so-called Golden Ratio = 1+
5
2
.
14According to www.dictionary.com, there are different values given for this conversion. We will stick with 33.6in3
for this problem.
4.3 Rational Inequalities and Applications 355
In Exercises 33 - 38, translate the following into mathematical equations.
33. At a constant pressure, the temperature T of an ideal gas is directly proportional to its volume
V . (This is Charless Law)
34. The frequency of a wave f is inversely proportional to the wavelength of the wave .
35. The density d of a material is directly proportional to the mass of the object m and inversely
proportional to its volume V .
36. The square of the orbital period of a planet P is directly proportional to the cube of the
semi-major axis of its orbit a. (This is Keplers Third Law of Planetary Motion )
37. The drag of an object traveling through a fluid D varies jointly with the density of the fluid
and the square of the velocity of the object .
38. Suppose two electric point charges, one with charge q and one with charge Q, are positioned r
units apart. The electrostatic force F exerted on the charges varies directly with the product
of the two charges and inversely with the square of the distance between the charges. (This
is Coulombs Law)
39. According to this webpage, the frequency f of a vibrating string is given by f =
1
2L
T
where T is the tension, is the linear mass15 of the string and L is the length of the vibrating
part of the string. Express this relationship using the language of variation.
40. According to the Centers for Disease Control and Prevention www.cdc.gov, a persons Body
Mass Index B is directly proportional to his weight W in pounds and inversely proportional
to the square of his height h in inches.
(a) Express this relationship as a mathematical equation.
(b) If a person who was 5 feet, 10 inches tall weighed 235 pounds had a Body Mass Index
of 33.7, what is the value of the constant of proportionality?
(c) Rewrite the mathematical equation found in part 40a to include the value of the constant
found in part 40b and then find your Body Mass Index.
41. We know that the circumference of a circle varies directly with its radius with 2 as the
constant of proportionality. (That is, we know C = 2r.) With the help of your classmates,
compile a list of other basic geometric relationships which can be seen as variations.
15Also known as the linear density. It is simply a measure of mass per unit length.
356 Rational Functions
4.3.3 Answers
1. x = 6
7
2. x = 1, x = 2 3. x = 1
4. x = 6, x = 2 5. No solution 6. x = 0, x = 2
2
7. (2,) 8. (2, 3]
9. (1, 0) (1,) 10. [0,)
11. (,3) (3, 2) (4,) 12.
(
3,1
3
)
(2, 3)
13. (1, 0] (2,) 14. (,3) (2,1) (1,)
15. (, 1] [2,) 16. (,6) (1, 2)
17. (,3)
[
2
2, 0
]
[
2
2, 3
)
18. No solution
19. [3, 0) (0, 4) [5,) 20.
(
1,1
2
]
(1,)
21. 4.5 miles per hour 22. 24 miles per hour 23. 3600 gallons
24. 40
3
13.33 minutes 25. 3 hours
26. The absolute minimum of y = C(x) occurs at (75.73, 59.57). Since x represents the number
of game systems, we check C(75) 59.58 and C(76) 59.57. Hence, to minimize the average
cost, 76 systems should be produced at an average cost of $59.57 per system.
27. The width (and depth) should be 10.00 centimeters, the height should be 5.00 centimeters.
The minimum surface area is 300.00 square centimeters.
28. The width of the base of the box should be 4.12 inches, the height of the box should be
6.67 inches, and the depth of the base of the box should be 5.09 inches; minimum surface
area 164.91 square inches.
29. The dimensions are 7 feet by 14 feet; minimum amount of fencing required 28 feet.
30. (a) V = r2h (b) S = 2r2 + 2rh
(c) S(r) = 2r2 + 67.2
r
, Domain r > 0 (d) r 1.749 in. and h 3.498 in.
31. The radius of the drum should be 1.05 feet and the height of the drum should be 2.12
feet. The minimum surface area of the drum is 20.93 cubic feet.
32. P(t) < 100 on (15, 30), and the portion of this which lies in the applied domain is [0, 30). Since t = 0 corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County. 4.3 Rational Inequalities and Applications 357 33. T = kV 34. 16 f = k 35. d = km V 36. P 2 = ka3 37. 17 D = k2 38. 18 F = kqQ r2 39. Rewriting f = 1 2L T as f = 1 2 T L we see that the frequency f varies directly with the square root of the tension and varies inversely with the length and the
