Express the revenue received for this charter flight as a function of the number of unsold seats.. c Determine the number of unsold seats that will result in the maximum revenue for the
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P R O B L E M S F O R S E C T I O N 6 4
1 A seat on a round-trip charter flight to Cairo costs $720 plus a surcharge of $10 for every unsold seat on the airplane (If there are 10 seats left unsold, the airline will charge each passenger $720 + $100 = $820 for the flight.) The plane seats 220 travelers and only round-trip tickets are sold on the charter flights
(a) Let x = the number of unsold seats on the flight Express the revenue received for
this charter flight as a function of the number of unsold seats (Hint: Revenue =
(price + surcharge)(number of people flying).)
(b) Graph the revenue function What, practically speaking, is the domain of the function?
(c) Determine the number of unsold seats that will result in the maximum revenue for the flight What is the maximum revenue for the flight?
2 Troy is interested in skunks and has purchased a bevy of them to study He plans to keep the skunks in a rectangular skunk corral, as shown below He will have a divider across the width in order to separate the males and females He has 510 meters of fencing to make his corral
(a) Let x = the width of the corral and y = the length of the corral Use the fact that Troy has only 510 meters of fencing to express y in terms of x
(b) Express A, the total area enclosed for the skunks, as a function of x
(c) Find the dimensions of the corral that maximize the total area inside the corral
divider
x
y
3 The height of a ball (in feet) t seconds after it is thrown is given by
h(t )= −16t2+ 32t + 48 = −16(t + 1)(t − 3)
(a) Graph h(t) for the values of t for which it makes sense Below it graph v(t) Be sure that v(t) looks like the derivative of h(t)
(b) From what height was the ball thrown?
(c) What was the ball’s initial velocity? Was it thrown up or down? How can you tell? (d) Was the ball’s height increasing or decreasing at time t = 2?
(e) At what time did the ball reach its maximum height? How high was it then? What was its velocity at that time?
(f ) How long was the ball in the air?
(g) What is the ball’s acceleration? Does this make physical sense?
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4 We know that Revenue = (price) · (quantity) Suppose a certain company has a mo-nopoly on a good If the company wants to increase its revenue it can do so by raising its prices up to a certain point However, at some point the price becomes so high that there are not enough buyers and the revenue actually goes down Therefore, if a mo-nopolist is attempting to maximize revenue, the momo-nopolist must look at the demand curve Suppose the demand curve for widgets is given by
p= 1000 − 4q, where p is measured in dollars and q in hundreds of items
(a) Express revenue as a function of price and determine the price that maximizes the monopolist’s revenue
(b) What price(s) gives half of the maximum revenue?
5 Suppose that q, the quantity of gas (in gallons) demanded for heating purposes, is given
by q = mp + b, where m and b are constants (m negative and b positive) and p is the price of gas per gallon The gas company is interested in its revenue function (a) Explain why m is negative
(b) Express revenue as a function of price (Note that m and b are constants, so if you express revenue in terms of m, b, and p, you have expressed revenue as a function
of price.) (c) Graph your revenue function, labeling all intercepts
(d) From your graph, determine the price that maximizes revenue (Your answer will
be in terms of m and b.) (e) Find Rand graph it
6 A catering company is making elegant fruit tarts for a huge college graduation cele-bration The caterer insists on high quality and will not accept shoddy-looking tarts When there are 7 pastry chefs in the kitchen they can each turn out an average of 44 tarts per hour The pastry kitchen is not very large; let us suppose that for each additional pastry chef put to the fruit-tart task the average number of tarts per chef decreases by
4 tarts per hour (Assume that reducing the number of chefs will increase the average production by 4 tarts per hour, until the number of chefs has decreased to 3 At that point reducing the number of chefs no longer increases the productivity of each chef.) (a) How many chefs will yield the optimum hourly fruit-tart production?
(b) What is the maximal hourly fruit-tart production?
(c) How many chefs are in the kitchen if the fruit-tart production is 320 tarts per hour?
7 The function R(p) = 35p(75 − p) gives revenue as a function of price, p, where the price is given in dollars
(a) Find the price at which the revenue is maximum
(b) What is the maximum price?
8 David and Ben are sitting in a tree house sorting through a basket of apples They find two that are rotten and decide to toss them into a garbage bin on the ground below The height of Ben’s apple t seconds after he throws it is given by
B(t )= −16t2+ 4t + 10,
Trang 36.4 The Free Fall of an Apple: A Quadratic Model 243 and the height of David’s apple t seconds after he throws it is given by
D(t )= −16t2− 2t + 10
(a) From what height are the apples thrown?
(b) What is the maximum height of Ben’s apple? Explain your answer
(c) What is the maximum height of David’s apple?
(d) Does Ben toss the apple up or throw it down? Does David toss the apple up or throw it down? Explain your reasoning
(e) How many seconds after he throws it does Ben’s apple hit the ground?
9 Amelia is a production potter If she prices her bowls at x dollars per bowl, then she can sell 120 − 5x bowls every week
(a) For each dollar she increases her price how many fewer bowls does she sell? (b) Express her weekly revenue as a function of the price she charges per bowl (c) Assuming that she can produce bowls more rapidly than people buy them, how much should she charge per bowl in order to maximize her weekly revenue? (d) What is her maximum weekly revenue from bowls?
Trang 5C H A P T E R
The Theoretical Backbone:
Limits and Continuity
AND A DEFINITION
The idea of taking a limit is at the heart of calculus The limiting process allows us to move from calculating average rates of change to determining an instantaneous rate of change; in other words, it allows us to determine the slope of a tangent line We will need it to tackle another fundamental problem in calculus, calculating the area under a curve We’ve also seen that the language of limits is useful as a descriptive tool
Roots of the limiting processes can be found in the work of the ancient Greek mathe-maticians Archimedes essentially used a limiting process in studying the area of a circle, calling his technique involving successive approximations “the method of exhaustion.” The limiting process was instrumental in the work of Isaac Newton and Gottfried Leibniz as they developed calculus and can also be found in work of their predecessors It is inter-esting that these early practitioners of calculus in the late 1600s achieved many useful and valid results without carefully defining the notion of a limit Not until substantially later were mathematicians able to put the work of their predecessors on solid ground with a rig-orous definition of a limit While we will not work much with the rigrig-orous definition, we will use this chapter to put the notion of limit on more solid ground
The challenge of computing instantaneous velocity from a displacement function, s(t ), led us to limits To find the instantaneous velocity at, say, t = 3, we took successive approximations using average velocity over the interval [3, 3 + h]
245
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(3, s(3))
(3 + h, s(3 + h))
∆t
∆ displacement
∆ time
∆s
s ′(3) ≈
s (3 + h) – s(3)
h
s ′(3) ≈
Figure 7.1
If h = 0 this expression is undefined; we have 00 But the closer h is to zero the better the approximation becomes We are interested in what happens tos(3+h)−s(3)h as h approaches zero but is not equal to zero This problem motivates our definition of limit
What Do We Mean by the Word “Limit”?
Below we give a short answer, which we will expand upon throughout the section limx →2f (x)= 5 means f (x) stays arbitrarily close to 5 provided that x is sufficiently close to 2, but not equal to 2
It does not tell us that f (2) = 5 It gives us no information about f (2); f (2) could be
5, or√
3, or undefined
However, we can guarantee that the difference between f (x) and 5 is smaller than any
positive number, no matter how miniscule, if x is close enough to 2 (but not equal to 2)
limx→∞f (x)= 6 means that the values of f (x) stay arbitrarily close to 6 provided x
is large enough
limx →2f (x)= ∞ means that f (x) increases without bound as x approaches 2 We’ll clarify the meaning of “arbitrarily close to” through the next two examples
EXAMPLE 7.1 Argue convincingly that if g(x) =x1, then limx →∞g(x)= 0
SOLUTION It “appears,” from the graph of g(x) =x1 (see Figure 7.2), that limx→∞1x = 0; numerical
evidence suggests this hypothesis as well
g
g (x) =
x
x
10 100 1000 10000
10 5
10 6
.1 01 001 0001
10 –5
10 –6
Figure 7.2
Trang 77.1 Investigating Limits—Methods of Inquiry and a Definition 247
But appearances alone can be deceiving For instance, suppose we graph h(x) =x1+ 10−15 Adding 10−15to g(x) shifts the graph of g(x) vertically up by 10−15 But 10−15is such a miniscule number that it is difficult to distinguish between h and g graphically (particularly using a graphing calculator) Yet if limx →∞ 1x= 0 then limx →∞(1x+ 10−15)ought to be
10−15 So although the graph in Figure 7.2 is very useful, it isn’t convincing evidence that limx→∞ 1x= 0
The table of values suggests a way of nailing things down Since 1x is positive and decreasing, we see that
g(x)is within 10−5of zero provided x > 105, g(x)is within 10−6of zero provided x > 106, g(x)is within 10−100of zero provided x > 10100
Even this last statement is not enough to show that limx→∞g(x)= 0 We must show that g(x)will be arbitrarily close to zero for all x large enough This means someone can issue
a challenge with any miniscule little number Let , (read: epsilon) be any small positive number; , can be excruciatingly small We must show that for x large enough, g(x) is within , of 0
g(x)=1
x is within , of zero provided x >1
, We’re done
In our next example we’ll look at limx →∞
1 2
x Before launching in, first we note that we define12nfor n a positive integer as 12 multiplied by itself n times:
1
2 ·1
2 ·1
2 .
1 2
n times
Using the rules of exponent algebra, which can be reviewed in either the Algebra Appendix
or Chapter 9 for any rational exponent x, we can define12xfor x any rational number For now we’ll deal with irrational exponents simply by approximating the irrational number
by a sequence of rational ones Suppose x is irrational, r and s are rational, and r < x < s Then12r<12x<12s We define bxso as to make the graph of bxcontinuous A more satisfactory definition can be given after taking up logarithmic functions
EXAMPLE 7.2 Argue convincingly that limx→∞12x= 0
SOLUTION Let’s begin by trying to get a feel for what happens to21x as x increases without bound
Consider the graph of f (x) =12x=21x on the following page
Trang 8248 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
f
x
1
Figure 7.3
A second way of getting a feel for the limit is to look numerically at the outputs of
f (x)=21x as x gets increasingly large
x f (x)=21x = (0.5)x(approximate values)
20 9.5 × 10−7= 0.00000095
50 8.8 × 10−16= 0.00000000000000088
100 7.8 × 10−31
200 6.2 × 10−61
300 4.9 × 10−91
It “appears” that limx→∞ 21x = 0 This is looking quite convincing; 4.9 × 10−91is quite close to zero But consider the following When asked to compute 0.5328, a TI-81 calculator gives approximately 1.8 × 10−99 But when asked for 0.5329, it gives the answer as 0 In fact, according to this calculator, for x > 329,21x = 0 We know this is false; a fraction can
be equal to zero only if its numerator is zero, and here the numerator is 1 It appears that the calculator rounds 10−100off to zero How can we be sure that limx →∞21x = 0, and not
10−120for instance? It is not quite good enough to simply say 0 < f (x) < 10−99provided
x >330.1
As in Example 7.1, we must show that f (x) will be arbitrarily close to zero provided
that x is large enough Again, a challenge is issued with any excruciatingly small number ,; we must show that f (x) is within , of 0 for x big enough To figure out how big is “big enough,” let’s compare f (x) =21x with g(x) =x1
1
Trang 97.1 Investigating Limits—Methods of Inquiry and a Definition 249
y
g (x) =
x
x
1
f (x) =
f (x) = 1x(approximated)
2
x
1 2
g (x) =
x x1
20 50 100 200
.05 02 01 005
9.5 × 10 –7
8.8 × 10 –16
7.8 × 10 –31
6.2 × 10 –61
Figure 7.4
Notice that21x <x1for all x > 0 (This is equivalent to the statement that if x > 0, then
2x> x.)
So for x > 0, if1x is within , of zero, then21x is certainly within , of zero
And21x is within , of zero provided x >1, (Notice by looking at the table of values that for large x, 21x is much, much smaller than 1x, so requiring that x >1, is overkill—but that’s all right.)
The last two examples highlight a couple of important ideas about limits in general and limx→∞f (x)= L in particular:
1.Graphical and numerical investigations are both useful methods of inquiry that can provide compelling data from which to arrive at a conjecture about a limit They cannot, however, be conclusive on their own
2.When we say “f (x) is arbitrarily close to L” we mean that the distance between f (x) and L can be made arbitrarily small To be arbitrarily small means that we can answer
a challenge set out by any miniscule positive number ,, no matter how excruciatingly
small , may be, that the distance between f (x) and L can be made less than , given certain conditions on x
EXAMPLE 7.3 Find limx→3x
2
and closer to 3 but is not equal to 3?” Intuitively, it should make sense that limx →3x2= 1.5 There is nothing particularly special happening to x2around x = 3
f
f (x) =
x
x
2
2 3
x x2
2.9 2.99 2.999 2.99999 3.00001 3.001 3.01 3.1
1.45 1.495 1.4995 1.499995 1.500005 1.5005 1.505 1.55 3
Figure 7.5
Trang 10250 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
More rigorously, we can show that if x is close enough to 3 (but not equal to 3), thenx2can
be made to stay arbitrarily close to 1.5 as follows
Look at Figure 7.6(a) We can see that if x is within 1 unit of 3, then f (x) is within 0.5 units of 1.5 Similarly, we see in Figure 7.6(b) that if x is within 0.1 of 3, then f (x) is within 0.05 =0.12 of 1.5
f
f (x) =
x
x
2
2 1
1 2
1
f
f (x) =
x
x
2
1.55 1.5 1.45
.1
2 = 05 05
2.9 3 3.1
Figure 7.6
Because f (x) is a straight line with slope 1/2, we can see that the ratio fx is always 0.5 We use this to ensure that the distance between f (x) and 1.5 can be made arbitrarily For , any positive number, no matter how excruciatingly small, if x is within 2, of 3
This last example illustrates what we mean by limx →3f (x)= L Showing that f (x) stays arbitrarily close to L for x close enough to 3 (but not equal to 3) is equivalent to the following: Given the challenge of any excruciatingly small positive ,, we can guarantee that f (x) will be within , of L provided x is close enough to 3 We can write this more compactly: If f (x) is within , of L, then the distance between f (x) and L is less than , Let’s use absolute values to express this distance
The distance between f (x) and L is |f (x) − L|
Similarly, the distance between x and 3 is |x − 3|
To show that x is within some distance δ (delta for distance) of 3 but not equal to 3, we write 0 < |x − 3| < δ
We arrive at the following definition, which we use in the examples which follow
D e f i n i t i o n
limx →3f (x)= L if, for every excruciatingly small positive number ,, we can come
up with a distance δ so that
|f (x) − L| < , provided 0 < |x − 3| < δ
This is the formal definition of a limit2 (where we can substitute “a” for 3 to get limx →af (x)= L) We applied this definition in the last example As promised at the
2 f must be defined on an open interval around 3, although not necessarily at x = 3.
... compactly: If f (x) is within , of L, then the distance between f (x) and L is less than , Let’s use absolute values to express this distanceThe distance between f (x) and L is |f (x) − L|
Similarly,... They cannot, however, be conclusive on their own
2.When we say “f (x) is arbitrarily close to L” we mean that the distance between f (x) and L can be made arbitrarily small To be... to this calculator, for x > 329,21x = We know this is false; a fraction can
be equal to zero only if its numerator is zero, and here the numerator