Chapter 4
Analyzing Change: Applications of Derivatives
Section 4.1 Approximating Change
1. 32% − (4 percentage points per hour)
(
( 13 hour ) = 30 23 %
)
1 hour = 316 2 mph
2. 300 mph + (200 mph per hour) 12
3
3. f (3.5) ≈ f (3) + f ′(3)(0.5) = 17 + 4.6(0.5) = 19.3
4. g (7.25) ≈ g (7) + g ′(7)(0.25)
= 4 + (−12.9)(0.25) = 0.775
5. a. Increasing production from 500 to 501 units will increase total cost by approximately $17.
b. If sales increase from 150 to 151 units, then profit will increase by approximately $4.75.
6. a. Increasing sales from 500 to 501 units will increase revenue by approximately $10.00 and
cost by approximately $13.00.
b. Increasing sales from 10 to 11 units will decrease profit by $3.46.
7. A marginal profit of –$4 per shirt means that at this point the fraternity’s profit is decreasing by
$4 for each additional shirt sold. The fraternity should consider selling fewer shirts or increasing
the sales price.
8. No. If the marginal profit at x is negative, then as x is increased the profit will decrease, but it
will not necessarily be negative.
9.
Using the two points (70,8000 and 54,0)
$8000
Slope of tangent line ≈
= $500 per year
16 years of age
of age
Annual premium
for 70-year-old ≈ $8000
Annual premium
 $500 
for 72-year-old ≈ $8000 + 
 (2 years) = $9000
 year 
(Estimates will vary.)
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141
142
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
10.
Slope of tangent line
10 years
≈
= 2.5 years per decade
4 decades
Life expectancy in 1990 ≈ 65 years
Life expectancy in 2000
≈ 65 years + (2.5 years/decade)(1 decade)
= 67.5 years
(Estimates will vary.)
11. a.
Slope of tangent line is approximately
97 billion dollars
= 97 billion dollars per billion dollars
1 billion dollars
(revenue dollars per sales dollars).
Revenue is approximately $614 billion when $6 billion is spent on advertising.
Revenue is approximately 614 + 0.5(97) = 662.5 billion dollars when $6.5 billion is spent on
advertising. (Discussion will vary.)
b. R(6.5) ≈ $658 billion
c. The solution from the model is more accurate than that which is derived from an
interpretation of the graph, because it is difficult to accurately draw a tangent line on so small
a graph.
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Section 4.1: Approximating Change 143
Calculus Concepts
12. a.
Slope of tangent line
−50 million kg
≈
4 years
= –12.5 million kg per year
CFC-11 releases in 1992 ≈ 170 million kg
CFC-11 releases in 1993 ≈ 170 million kg + (–12.5 million kg per year)(1 year)
= 157.5 million kg
b. C(5) ≈ 165 million kg
c. The tangent line estimate is closer to the actual amount, because it is an underestimate of the
model value, and the model overestimates the actual value.
13. a.
P ( x) = 268.79(1.013087 x ) thousand people in year x
P ' ( x) = 268.79(ln 1.013087)(1.013087 x ) thousand people per year in year x
In 2000 the population of South Carolina was increasing by 53.6 thousand people per year.
b. Between 2000 and 2003, the population increased by approximately 160.8 thousand people.
c. By finding the slope of the tangent line at 2000 and multiplying by 3, we determine the
change in the tangent line from 2000 through 2003 and use that change to estimate the change
in the population function.
14. a. A(t) = 120(1.126t) thousand dollars. A’(t) = 120(ln 1.126)(1.126t) thousand dollars/year.
b. A’(10) = 46.656. The investment, at ten years, is changing at a rate of $46,656/year.
c. Growth in the 1st half of the 11th year will be ≈
1
2
(46.656) or $23,328.
A'( t )
A'(10)
d. The percentage rate of change after 10 years is A( t ) i100% = A(10) i100% =
11.9%/year.
e. The percentage rate of change and the perdentage change, while both constants, are not equal.
The percentage change is (b-1) i 100% = 12.6%. The percentage rate of change is the log of
b i 100%= 11.9%.
15. a. The population was growing at a rate of 2.52 million people per year in 1998.
b. Between 1998 and 1999, the population of Mexico increased by approximately
2.52 million people.
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
16. a. G (t ) = −0.044t 3 + 0.92t 2 + 38 points after t hours of study
G (11) ≈ 90.8 points
G ′(t ) = −0.132t 2 + 1.84t
i. G ′(11) ≈ 4.27 points per hour of study
ii. G(12) ≈ 94.4 points
b. G (12) = G (11 + 1) ≈ G (11) + G ′(11)(1)
≈ (90.76 points) + (4.27 points per hour)(1 hour) = 95.03 points
This is an overestimate, because the graph of the model is concave down for values of t
between 11 and 12.
17. a. In 1998 the amount was increasing by 1.15 million pieces per year.
b. We would expect an increase of approximately 1.15 million pieces between 1998 and 1999.
c. p(24) – p(23) ≈ 1.3 million pieces
d. 101.9 – 100.4 = 1.5 million pieces
e. As long as the data in part d were correctly reported, the answer to part d is the most accurate
one.
18. a. Production Costs = C ( p ) = 0.16 p 3 − 8.7 p 2 + 172 p + 69.4 dollars when p units are produced
hourly.
b. C ′( p ) = 0.48 p 2 + 17.4 p + 172 dollars per unit when p units are produced hour.
C ′(5) = 97 dollars per unit
C ′(20) = 16 dollars per unit
C ′(30) = 82 dollars per unit
When 5 units are produced hourly, the hourly cost is increasing at a rate of $97 per additional
unit produced in an hour. When 20 units are produced hourly, the hourly cost is increasing at
a rate of $16 per additional unit produced in an hour. When 30 units are produced hourly, the
hourly cost is increasing at a rate of $82 per additional unit produced in an hour.
c. Cost of 6th unit: C (6) − C (5) ≈ $822.76 − $731.90 = $90.86
Cost of 21st unit: C (21) − C (20) ≈ $1326.46 − $1309.40 ≈ $17.06
Cost of 31st unit: C (31) − C (30) ≈ $1807.26 − $1719.40 = $87.86
d. The model is concave down at x = 5, but it is concave up at x = 20 and x = 30.
d.
C ( p)
= 0.16 p 2 − 8.7 p + 172 p + 69.4 p −1 dollars per unit
p
when p units are produced hourly
A( p ) =
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Section 4.1: Approximating Change 145
Calculus Concepts
e.
A′( p ) = 0.32 p − 8.7 + 172 p −1 + 69.4 p −2 dollars per unit per unit
when p units are produced hourly
When the production level is at 5, the average cost is changing at a rate of
A′(5) ≈ 30.08 dollars per unit per unit produced hourly.
When the production level is at 20, the average cost is changing at a rate of
A′(20) ≈ −6.47 dollars per unit per unit produced hourly.
When the production level is at 30, the average cost is changing at a rate of
A′(30) ≈ 6.71 dollars per unit per unit produced hourly.
When 5 units are produced hourly, the per-unit cost increases by $30.08 per additional unit
produced hourly. When 20 units are produced hourly, the per-unit cost decreases by $6.47 per
additional unit produced hourly. When 30 units are produced hourly, the per-unit cost
decreases by –$6.71 per additional unit produced hourly.
19. a.
A = 300(1 +
.065 (12 t )
12
)
b. A = 300(1.06697)t
c. A(2) = $341.53
d. A’(2) = 300(ln 1.06697)(1.06697)2 = 22.14 dollars/year.
e. A(2.25) ≅ A(2) + .25 A '(2) = $347.07
20. a.
A = 2000(1 +
.032 (12 t )
12
)
b. A = 2000(1.0325)t
c. A(5) = $2346.52
d. A’(5) = 2000(ln 1.0325)(1.0325)5 = 74.99 dollars/year.
e. A(5.5) ≅ A(5) + .5 A '(5) = $2384.02
21. Sales
(
)
a. R ( x) = −7.032 ⋅ 10−4 x 2 + 1.666 x + 47.130 dollars when x hot dogs are sold, 100 < x < 1500.
b. Cost: C ( x ) = 0.5 x dollars when x hot dogs are sold, 100 < x < 1500.
Profit: P ( x ) = R ( x ) − C ( x ) = ( −7.032 ⋅ 10 −4 ) x 2 + 1.166 x + 47.130 dollars when x hot
dogs are sold, 100 < x < 1500.
c. Marginal Revenue = R’(x) = -.0014x + 1.666 dollars/hot dog.
x
(hot dogs)
200
800
1100
1400
R′( x )
c ′( x )
p′( x )
(dollars per hot dog)
1.38
0.54
0.12
(dollars per hot dog)
0.50
0.50
0.50
0.50
(dollars per hot dog)
0.88
0.04
−0.38
−0.80
−0.30
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
If the number of hot dogs sold increases from 200 to 201, the revenue increases by
approximately $1.38 and the profit increases by approximately $0.88. If the number increases
from 800 to 801, the revenue increases by 0.54, but the profit sees almost no increase (4
cents). If the number increases from 1100 to 1100, the increase in revenue is only
approximately 12 cents. Because this marginal revenue is less than the marginal cost at a
sales level of 1100, the result of the sales increase from 1100 to 1101 is a decrease of $0.38 in
profit. If the number of hot dogs increases from 1400 to 1401, revenue declines by
approximately 30 cents and profit declines by approximately 80 cents.
The marginal values in part c are the
slopes of the graphs shown here. For
example, at x = 800, the slope of the
revenue graph is $0.54 per hot dog, the
slope of the cost graph is $0.50 per hot
dog, and the slope of the profit graph is
$0.04 per hot dog. We see from the graph
that maximum profit is realized when
approximately 800 hot dogs are sold.
Revenue is greatest near x = 1100,
so the marginal revenue there is small.
However, once costs are factored in, the
profit is actually declining at this sales
level. This is illustrated by the graph.
d.
22. Production
a. C ( x) = 0.068 x3 − 2.933x 2 + 55.269 x + 146.983 dollars
when x hundred balls are produced hourly
b. C ′( x) = 0.204 x 2 − 5.865 x + 55.269 dollars per hundred balls
when x hundred balls are produced hourly
C ′(10) ≈ 16.99 dollars per hundred balls or $0.17 per ball
The hourly cost will increase by approximately $0.17 for each additional ball produced in an
hour.
c.
d.
e.
C ′(3) ≈ 39.51 dollars per hundred balls or $0.395 per ball
C ′(21) ≈ $21.94 dollars per hundred balls or $0.22 per ball
When 300 balls are produced hourly, the hourly cost will increase by approximately $0.395
for each additional ball produced in an hour. When 2100 balls are produced hourly, the
hourly cost will increase by approximately $0.22 for each additional ball produced in an hour.
C ( x)
= 0.068 x 2 − 2.933 x + 55.269 + 149.983 x −1 dollars per hundred balls
x
when x hundred balls are produced hourly
A( x) =
A′( x) = 0.136 x − 2.933 − 146.983 x −2 dollars per hundred balls per hundred balls
when x hundred balls are produced hourly
A′(3) ≈ −18.86 dollars per hundred balls per additional hundred balls produced hourly or
–$0.00189 per ball per additional ball produced hourly
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Calculus Concepts
Section 4.1: Approximating Change 147
A′(17) ≈ −1.13 dollars per hundred balls per additional hundred balls produced hourly or
–$0.000113 per ball per additional ball produced hourly
When 300 balls are produced hourly, the cost per ball will decrease by approximately
$0.00189 for each additional ball produced in an hour. When 1700 balls are produced hourly,
the cost per ball will decrease by approximately $0.000113 for each additional ball produced
in an hour.
23. CPI
United States:
a. A(t ) = 0.109t 3 − 1.555t 2 + 10.927t + 100.320 t years after 1980
b.
A′(t ) = 0.327t 2 − 3.111t + 10.927 index points per year t years after 1980
A′(7) ≈ 5.2 index points per year
c. 1988 CPI estimate:
(CPI in 1987) + A′(7) (1 year) ≈ 137.9 + (5.2 index points per year)(1 year)
= 143.1
Note: The estimate can also be calculated using the value of A(7) instead of the actual CPI in
1987. Because the model closely agrees with the actual value in 1987, the value of this
estimate is not significantly affected by this choice.
Canada:
a. C (t ) = 0.150t 3 − 2.171t 2 + 15.814t + 99.650 t years after 1980
b. C ′(t ) = 0.450t 2 − 4.343t + 15.814 index points per year t years after 1980
C ′(7) ≈ 7.5 index points per year
c. 1988 CPI estimate:
(CPI in 1987) + C ′(7) (1 year) ≈ 155.4 + (7.5 index points per year)(1 year)
= 162.9
Note: The estimate can also be calculated using the value of C(7) instead of the actual CPI in
1987. Because the model closely agrees with the actual CPI in 1987, the value of this
estimate is not significantly affected by this choice.
Peru:
a. P (t ) = 85.112(2.013252t ) t years after 1980
b.
P′(t ) = 85.112(ln 2.013252)(2.013252t )
≈ 59.558(2.013252t ) index points per year t years after 1980
P′(7) ≈ 7984 index points per year
c. 1988 CPI estimate:
(CPI in 1987) + P′(7) (1 year) ≈ 11,150 + (7984 index points per year)(1 year)
= 19,134
Note: The estimate can also be calculated using the value of P(7) instead of the actual CPI in
1987. If this is done, the estimate will be approximately 19,394.
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
Brazil:
a. B (t ) = 73, 430(2.615939t ) t years after 1980
b.
B′(t ) = 73.430(ln 2.615939)(2.615939t )
≈ 70.612(2.615939t ) index points per year t years after 1980
B′(7) ≈ 59,193 index points per year
c. 1988 CPI estimate:
(CPI in 1987) + B′(7) (1 year) ≈ 77,258 + (59,193 index points per year)(1 year)
= 136,451
Note: The estimate can also be calculated using the value of B(7) instead of the actual CPI in
1987. If this is done, the estimate will be approximately 120,748.
24. Revenue
a. Revenue = R ( x ) = −12.16 x 2 + 254.28 x − 105.60 dollars, where x = the price for a large
one-topping pizza, in dollars. 9.25 < x < 14.25.
b. R '( x ) = −24.32 x + 254.28 dollars per dollar of pizza price.
R’(9.25) = $29.32 When the price of a large one-topping pizza is $9.25, the revenue is
increasing by $29.32 for every additional dollar in the price of pizza.
c. ∆Revenue = (10.25 − 9.25)(29.32) = $29.32
d. R’(11.50) = -$25.4
When the price of a large one-topping pizza is $11.50, the revenue is
decreasing by $25.40 for every additional dollar in the price of pizza.
e. ∆Revenue = (11.50 − 12.50)( −25.4) = −$25.40
f. In both cases the graph is concave down.
25. Advertising
Note: This Activity can be solved using either a cubic model or a logistic model. The following
solution uses a cubic model.
a.
R ( A) = −0.158 A3 + 5.235 A2 − 23.056 A + 154.884 thousand dollars of revenue when A
thousand dollars is spent on advertising. 5 < A < 19.
b.
R′( A) = −0.473 A3 + 10.471A2 − 23.056 thousand dollars of revenue per thousand dollars of
advertising when A thousand dollars is spent on advertising
R′(10) ≈ 34.3 thousand dollars of revenue per thousand dollars of advertising
When $10,000 is spent on advertising, revenue is increasing by $34.3 thousand per thousand
advertising dollars. If advertising is increased from $10,000 to $11,000, the car dealership
can expect an approximate increase in revenue of $34,300.
c.
R′(18) ≈ 12.0 thousand dollars of revenue per thousand of dollars of advertising
When $18,000 is spent on advertising, revenue is increasing by $12.0 thousand per thousand
advertising dollars. If advertising is increased from $18,000 to $19,000, the car dealership
can expect an approximate increase in revenue of $12,000.
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Section 4.1: Approximating Change 149
Calculus Concepts
26. Newspapers
a. n( x) = 0.008 x 3 − 0.321x 2 + 3.457 x + 51.588 million newspapers x years after 1980. The
curvature of the scatter plot suggests either a logistic or a cubic function, except that the data
point for 1986 is lower than that for 1988. A cubic would be the better model for the
max/min behavior.
b. n(27) ≈ 67.1 million newspapers
c.
n' ( x) = 0.024 x 2 − 0.642 x + 3.457 million newspapers per year
n’(18) ≈ -0.4 million newspapers per year
d. n’(20) ≈ 0.1 million newspapers per year. We approximate the change in circulation between
1990 and 1991 to be an increase of 0.1 million newspapers.
27. One possible answer: Close to the point of tangency, a tangent line and a curve are close to one
another. The farther away from the point of tangency we move, the more the tangent line deviates
from the curve. Thus the tangent line near the point of tangency will usually produce a good
estimate, but the tangent line farther away from the point of tangency will produce a poor
estimate.
28. One possible answer: Because a line tangent to a point on a concave-up portion of a curve lies
below the curve near the point of tangency, estimates taken from it will be under approximations
(unless the tangent line cuts through the curve at some point). Because a line tangent to a point on
a concave-down portion of a curve lies above the curve near the point of tangency, estimates
taken from it will be over approximations (unless the tangent line cuts through the curve at some
point).
29. One possible answer: By definition f '( x) = lim
h →0
f ( x + h) − f ( x)
. Assuming h is relatively close
h
f ( x + h) − f ( x )
. Multiplying both sides of this approximation by h yields
to zero, f '( x) ≈
h
h ⋅ f '( x) ≈ f ( x + h) − f ( x) .
Section 4.2 Relative and Absolute Extreme Points
1. Quadratic, cubic, and many product, quotient, and composite functions could have relative
maxima or minima.
2. A graph of the function can be used to find the approximate values of the relative maxima and
minima. (If technology is used, very accurate approximations can be obtained.) The exact
values can be obtained by determining the exact output values where the derivative is zero,
provided that the derivative graph crosses (not just touches) the input axis at that value.
Additional relative minima or maxima may occur at the breakpoints of a piecewise continuous
function.
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
3.
The derivative is zero at the absolute maximum point.
4.
The derivative is zero at the relative minimum point and at the relative maximum point.
5.
The derivative is zero at the absolute maximum point marked with an X. The derivative where
the graph is broken, at the relative minimum, is undefined.
6.
The derivative is undefined at the absolute maximum point.
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Calculus Concepts
Section 4.2: Relative and Absolute Extreme Points 151
7.
The derivative is zero at both absolute maximum points. The derivative does not exist at the
relative minimum point.
8.
The derivative is undefined at the relative minimum point.
3
9. One possible answer: One such graph is y = x , which does not have a relative minimum or
maximum at x = 0 even though the derivative is zero at this point.
10. One possible answer: One such graph is shown to the right.
11. a. All statements are true.
b. The derivative does not exist at
x = 2 because f is not continuous there, so the third statement is false.
c. The slope of the graph is negative, f ′( x) < 0, to the left of x = 2 because the graph is
decreasing, so the second statement is false.
d. The derivative does not exist at
x = 2 because f is not smooth there, so the third statement is not true.
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
12.a. The slope of this graph is always positive, so the second statement is not true.
b. All statements are true.
c. All statements are true.
d. The third statement is not true because the derivative does not exist at x = 2.
13. One possible graph:
14.
One possible graph:
15. One possible graph:
16.
One possible graph:
17. a. The derivative formula is f’(x) = 2x + 2.5
b. Using technology, the relative minimum value is approximately -7.5625, which occurs at
x ≈ -1.25.
c.
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Section 4.2: Relative and Absolute Extreme Points 153
Calculus Concepts
18. a. g’(x) = -6x + 14.1
b. The relative maximum is 0.3675, which occurs at x = 2.35
19. a. h’(x) = 3x2 – 16x - 6
b. The relative maximum is 1.077, which occurs at x ≈ -0.352;
The relative minimum is -108.929, which occurs at x ≈ 5.685
20. a. j’(x) = 0.9x2 + 2.4x - 6
b. The relative maximum is 28.146, which occurs at x ≈ -4.239;
The relative minimum is -1.301, which occurs at x ≈ 1.573
21. a. f’(t) = 12(ln 1.5)(1.5t) + 12(ln 0.5)(0.5t)
b. This function is always increasing, so does not have a relative maximum
nor relative minimum.
22. a. j’(t) = -5e-t +
1
t
b. The relative maximum is 2.508 which occurs at t ≈ 0.259
23. a. g’(x) = .12x2 – 1.76x + 4.81
b. The relative maximum is 19.888, which occurs at x ≈ 3.633;
The relative minimum is 11.779, which occurs at x ≈ 11.034
c. On the closed interval [ 0 , 14.5 ],
The absolute minimum is found at the point (11.034, 11.779);
The absolute maximum is found at the point (3.633, 19.888)
24. a. The relative maximum is 2.286, which occurs at x ≈ 0.251;
There is no relative minimum.
b. On the closed interval [ 0 , 10 ],
the absolute minimum is found at the point (10, -9.230);
the absolute maximum is found at the point (.251, 2.286)
c. The graph of the derivative crosses the x-axis at the point x = 0.251 and no other time.
25. Grasshoppers
a. At 9.449 °C , the greatest percentage of eggs, 95.598%, eggs hatch.
b. 9.449 °C corresponds to 49 °F .
26. Population
The absolute minimum is found at the point (3.754, 6.156), which corresponds to 2004.
The absolute maximum is found at the point (27.116, 10.462), which corresponds to 2028.
27. River Rate
a. The flow rate for h = 0 was 123.02 cfs; for h = 11 it was 331.305 cfs.
b. The absolute minimum is found at the point (.388, 121.311), or when h ≈ 0.4 hours
The absolute maximum is found at the point (8.900, 387.975) or h ≈ 8.9 hours
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Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
28. Lake Level
Because 1996 was a leap year, the number of days from October 1, 1995, to July 31, 1996, was
31 + 30 + 31 + 31 + 29 + 31 + 30 + 31 + 30 + 31 = 305 days. We using technology to solve the
equation L′(d ) = 3(−5.345 ⋅ 10−7 )d 2 + 2(2.543 ⋅ 10−4 )d − 0.0192 = 0 , and we obtain the solutions
d ≈ 43.8, 273.4 . To find the maximum height of the lake for 0 ≤ x ≤ 305, we compare the
outputs at the endpoints with the outputs corresponding to the places where the derivative is 0:
(0, 6226.192)
(43.8, 6225.794)
(273.4, 6229.028)
(305, 6228.827)
The highest lake level was 6229.028 feet above sea level after approximately 273.4 days, which
is below 6229.1 feet above sea level. Yes, the lake remained below the maximum level.
29. Swim Time
a. S ( x) = 0.181x 2 − 8.463x + 147.376 seconds at age x years.
b. The model gives a minimum time of 48.5 seconds occurring at 23.4 years.
c. The minimum time in the table is 49 seconds, which occurs at 24 years of age.
30. Costs
a. Hourly Cost = C(x) = 0.0198x3 – 1.779x2 + 58.422x + 152.079 dollars,
when the production level is x units per hour, (1 < x < 61).
b. The marginal cost function is C’(x) = 0.0594x2 – 3.558x + 58.422 dollars.
At the production level of 40 units per hour, the marginal cost is $11.14 dollars per unit.
c. Average Hourly Cost = A(x) = 0.0198x2 – 1.779x + 58.422 + 152.079x-1 dollars,
when the production level is x units per hour, (1 < x < 61).
d. Using technology, the minimum average hourly cost occurs at (46.692, 21.783) the point
where 46.692 units per hour are being produced. The average hourly cost is $21.78, and
the total cost is A(46.692) = $1017.08.
31. Sales
a. A quadratic or exponential model can be used to model the data, but the exponential model
may be a better choice because it does not predict that demand will increase for prices
above $40. An exponential model for the data is R ( p ) = 316.765(0.949 p ) dozen roses
when the price per dozen is p dollars.
b. Multiply R(p) by the price, p. The consumer expenditure is E ( p ) = 316.765 p (0.949 p )
dollars spent on roses each week when the price per dozen is p dollars.
c. Using technology, we find that E(p) is maximized at p ≈ 19.16 dollars. A price of $19.16
per dozen maximizes consumer expenditure.
d. Profit is given by F ( p ) = E ( p ) − 6 R ( p ) = 316.765( p − 6)(0.949 p ).
Using technology, F(p) is maximized at p ≈ 25.16 dollars. A price of $25.16 per dozen
maximizes profit.
e. Marginal values are with respect to the number of units sold or produced. In this activity,
the input is price, so derivatives are with respect to price and are not marginals.
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Calculus Concepts
32.
Section 4.2: Relative and Absolute Extreme Points 155
Demand
a. A quadratic model for the data is p( x ) = 0.0866 x 2 − 10.125 x + 299.710 dollars when the
demand is x, that is, when x tenants desire the facilities. (Note that an exponential model
does not fit the data well. This is a rare case in which shifting the data up actually produces
a better-fitting exponential model.)
b. To find the revenue, multiply the price, p(x), by the demand, x.
R ( x) = 0.0866 x3 − 10.125 x 2 + 299.710 x dollars when the demand is x tenants
c. Examining a graph of R, we see that the maximum occurs between x = 15 and x = 25. On
this interval, the derivative is zero at x ≈ 19.86. Because the number of tenants must be an
integer, we compare the values of R(x) for x = 20 and x = 19. The greater output value
occurs at x = 20. The price corresponding to this demand is p(20) ≈ 131.85. Thus the
revenue is maximized at a price of approximately $132 and a demand of 20 tenants. The
marginal revenue at the maximum point is near zero.
33. Refuse
a.
G (t ) = 0.008t 3 − 0.347t 2 + 6.108t + 79.690 million tons of garbage taken to a landfill
t years after 1975
b. G ′(t ) = 0.025t 2 − 0.693t + 6.108 million tons of garbage per year t years after 1975
c. In 2005 the amount of garbage was increasing by G ′(30) ≈ 8.1 million tons per year.
Because the derivative graph exists for all input
values and never crosses the horizontal axis, G(t) has
no relative maxima.
d.
34. Price
a. Exponential model: A( p ) = 568.074(0.965582 p ) tickets sold on average when the price is p
dollars
Quadratic model: A( p ) = 0.15 p 2 − 16.007 p + 543.286 tickets sold on average when the
price is p dollars
The exponential model probably better reflects the probable attendance if the price is raised
beyond $35 because attendance is likely to continue to decline. (The quadratic model
predicts that attendance will begin to increase around $53.)
b. Multiply the exponential ticket function by the price, p, to obtain the revenue function.
R ( p ) = 568.074 p (0.965582 p ) dollars of revenue when the ticket price is p dollars.
c. Using technology, the maximum point on the revenue graph is approximately (28.55,
5966.86). This corresponds to a ticket price of $28.55, which results in revenue of
approximately $5967. The resulting average attendance is A(28.55) ≈ 209 .
Copyright © Houghton Mifflin Company. All rights reserved.
156
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
35. One possible answer: The graph shown below indicates there is an absolute maximum to the
right of -3 and an absolute minimum to the left of 1. A view of the graph showing more of the
horizontal axis indicates that y = 2 is a horizontal asymptote for the graph.
Use technology to find the absolute extrema, or solve the equation
y′ =
−2 x(2 x 2 − x + 3)
2
+
4x −1
=0
( x + 2)
x2 + 2
In either case you should find the absolute minimum point of approximately (0.732, 1.317) and
the absolute maximum point of approximately (−2.732, 2.183). Thus the absolute maximum is
approximately 2.18, and the absolute minimum is approximately 1.32.
2
36. One possible answer: The derivative of y is zero for three values of x: x = −3.5, x ≈ −1.049,
and x ≈ 1.549. A graph indicates that x = −3.5 and x ≈ 1.549 correspond to relative minima and
x ≈ −1.049 correspond- ing to a relative maximum. There are no places where the derivative is
not defined. Observing the end behavior of the graph (rising infinitely on both sides) and
comparing the values of y for
x = −3.5 and x ≈ 1.549, we conclude that there is no absolute maximum and the absolute
[
]
. ) + (1549
. ) 2 ( 35
. + 1549
. ) 2 ≈ −6.312.
minimum is y = 2 − 3(1549
Section 4.3 Inflection Points
1. Production
a. One visual estimate of the inflection points is (1982, 25) and (2018, 25). Note: There are
also “smaller” inflection points at approximately (1921, 2.5), (1927, 2), (1930, 2), and
(1935, 3).
b. The input values of the inflection points are the years in which the rate of crude oil
production is estimated to be increasing and decreasing most rapidly. We estimate that the
rate of production was increasing most rapidly in 1982, when production was
approximately 25 billion barrels per year, and that it will be decreasing most rapidly in
2018, when production is estimated to be approximately 25 billion barrels per year.
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 4.3: Inflection Points 157
2. Advertising
a.
b. The inflection point occurs when the rate at which revenue is increasing with respect to the
amount spent on advertising is greatest. It can be regarded as the point of diminishing
returns.
c.
Answers may vary. It is important to note that additional dollars spent after the inflection
point are generating diminishing returns.
3. For polynomial functions, as these appear to be, you can identify the function and its derivative
by noticing the number of inflection points. Because a derivative has a power one less than the
original function, it will also have one less inflection point. Thus graph b with two inflection
points is the function. Graph a with one inflection point is the derivative, and graph c with no
inflection points is the second derivative.
4. Using the same reasoning as in Activity 3, we conclude that graph a is the function, graph c is
the derivative, and graph b is the second derivative.
5. Graph c appears to have a minimum at −1 and an inflection point at −2. Graph b crosses the
horizontal axis at −1 and graph a crosses it at −2. Thus graph c is the function, graph b is the
derivative, and graph a is the second derivative.
6. Graph a appears to have a minimum at 0, a maximum at −2, and inflection points at −3 and
between −1 and 0. Graph b crosses the horizontal axis at −2 and 0 and graph c crosses it at
−3.4 and between −1 and 0. Thus graph a is the function, graph b is the derivative, and graph c
is the second derivative.
7. f’(x) = -3; f”(x) = 0
8. g’(t) = et; g”(t) = et
9. c’(u) = 6u – 7; c”(u) = 6
10. k’(t) = -4.2t + 7; k”(t) = -4.2
11. p’(u) = -6.3u2 + 7u; p”(u) = -12.6u + 7
12. f’(s) = 96s2 – 4.2s +7; f”(s) = 192s – 4.2
13. g’(t) = 37(ln 1.05)(1.05t); g”(t) = 37(ln 1.05)2(1.05t)
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158
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
14. h’(t) = -3(ln .02)(.02t); h”(t) = -3(ln .02)2(.02t)
15. f’(x) = 3.2x-1; f”(x) = -3.2x-2
16. g’(x) = 3e3x – x-1; g”(x) = -9e3x – x-2
17. L '(t ) = −131.04e
3.9t
(1 + 2.1e
3.9t −2
)
; L”(t) = L "(t ) =
18. L '(t ) = 199.2e −0.02t (1 + 99.6e −0.02t ) −2 ; L "(t ) =
−2146.4352e
7.8 t
(1 + 2.1e )
3.9 t
7.936e
3
+
−.04 t
+
511.056e
3.9 t
(1 + 2.1e )
3.9 t
−3.984e
2
−.02 t
(1 + 99.6e ) (1 + 99.6e )
−.02 t
3
−.02 t
2
19. f '( x) = 3x 2 − 12 x + 2 ; f "( x) = 6 x − 12 = 0 at x = 2
20. g '(t ) = -0.3t 2 + 2.4t + 3.6 ; g "(t ) = -0.6t + 2.4 = 0 at t = 4
21. Using technology, the second derivative = 0 at the point x = 3.356
22. Using technology, the second derivative = 0 at the point t = -17.918
23. There is no value for t for which the second derivative is zero, therefore there is no inflection
point.
24. There is no value for t for which the second derivative is zero, therefore there is no inflection
point.
25. a.
g ( x) = 0.04 x3 − 0.88 x 2 + 4.81x + 12.11
g ′( x) = 0.12 x 2 − 1.76 x + 4.81
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Section 4.3: Inflection Points 159
Calculus Concepts
b.
g ′′( x ) = 0.24 x − 176
.
The inflection point on the graph of g is
approximately (7.333, 15.834). This is a
point of most rapid decline.
26. a.
LABe− Bx
20
f ( x) =
. Using the formula
1 + 19e − 0.5x
f ′( x) =
(
20(1)(0.5) 19e−0.5 x
(1 + 19e−0.5x )
(
2
(1 + Ae− Bx ) 2
for the derivative, we have
) = 190e−0.5x 1 + 19e−0.5x −2
(
)
)(
(
)
)
−2 
d
 d − 0.5 x 
− 0.5 x −2
f ′′ ( x ) = 190
e
+ 190e − 0.5x 
1 + 19e − 0.5x
 1 + 19e

 dx

 dx

(
)
(

= 190  e−0.5 x (−0.5) 1 + 19e−0.5 x

(
= −95e − 0.5x 1 + 19e − 0.5x
)
−2
)
−2
(
+ e−0.5 x (−2) 1 + 19e −0.5 x
(
+ 3610e − x 1 + 19e − 0.5x
Copyright © Houghton Mifflin Company. All rights reserved.
)
−3
) (19e−0.5x ) (−0.5)
−3
160
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
b. The inflection point of f is approximately (5.889, 10). This is a point of most rapid increase.
27. Study Time
a.
P(t ) =
45
1 + 5.94e − 0.969125t
(
= 45 1 + 5.94e − 0.969125t
(
P ′(t ) = 45( −1) 1 + 5.94e − 0.969125t
(
) (5.94e
−2
−1
percent after studying for t hours
− 0.969125t
= 259.0471125e −0.969125t 1 + 5.94e −0.969125t
(
)
)
−2
)(−0.969125)
percentage points per hour
after studying for t hours
)(
)
 d − 0.969125t 
− 0.969125t −2
P ′′ (t ) = 259.0471125
e
 1 + 5.94e
 dx

(
)
−2 
d
1 + 5.94e − 0.969125t

+ 259.0471125e −0.969125t 
 dx

(
)
(
= 259.0471125 e − 0.969125t ( −0.969125) 1 + 5.94e − 0.969125t
(
− 0.969125t
+ 259.0471125e −0.969125t ( −2) 1 + 5.94e
)
) (5.94e
−3
−2
− 0.969125t
)(−0.969125)
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Section 4.3: Inflection Points 161
Calculus Concepts
(
≈ −251.049033e − 0.969125t 1 + 5.94e − 0.969125t
)
−2
+ 2982.462511e –1.93825t (1 + 5.94e –0.969125t ) –3 percentage points per hour
per hour after studying for t hours
.
. The inflection point on P is approximately
Solving P ′′ (t ) = 0 for t gives t ≈ 1838
(1.838, 22.5). After approximately 1.8 hours of study (1 hour and 50 minutes), the
percentage of new material being retained is increasing most rapidly. At that time,
approximately 22.5% of the material has been retained.
b. The answer agrees with that given in the discussion at the end of the section.
28. Population
a. p ′ ( x ) = 4(1.619 ⋅ 10 −5 ) x 3 − 3(1.675 ⋅ 10 −3 ) x 2 + 2(0.050) x − 0.308 percentage points per
year x years after the end of 2000
p′′( x) = 12(1.619 ⋅ 10−5 ) x 2 − 6(1.675 ⋅ 10−3 ) x + 2(0.050) percentage points per year per year
x years after the end of 2000
Solving for x in the equation p′′( x) = 0 gives x ≈ 13.44, 38.29 . The lower value
corresponds to a maximum slope value. To make sure it is the absolute maximum for the
years between 2000 and 2050, we compare the slope value at x ≈ 13.44 with the slope
values at the endpoints: p′(0) = −0.308 and p′(50) ≈ 0.225 . Both of these values are less
than the slope at x ≈ 13.44 . The percentage will be increasing most rapidly in 2014, when
the percentage will be approximately 8.05% and will be increasing at a rate of 0.2855
percentage point per year.
b. The second solution found in part a, x ≈ 38.29 , corresponds to a minimum on the
derivative graph. Again, we compare the slope at that inflection point with the slopes at the
endpoints to determine the year in which the most rapid decrease is expected to take place.
We find that the slope in 2000 has a greater magnitude than the slope for x ≈ 38.29 . Thus
the percentage will be decreasing most rapidly in 2000, when the percentage will be
approximately 6.69% and will be decreasing at a rate of 0.308 percentage point per year.
29. Grasshoppers
a. P (t ) = −0.00645t 4 + 0.488t 3 − 12.991t 2 + 136.560t − 395.154 percent
when the temperature is t°C
P′(t ) = −0.0258t 3 + 1.464t 2 − 25.982t + 136.560 percentage points per °C
when the temperature is t°C
P′′(t ) = −0.0774t 2 + 2.928t − 25.982 percentage points per °C per °C
when the temperature is t°C
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162
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
b. Because the graph of P′′ crosses the
t-axis twice, there are two inflection
points. These are approximately
(14.2, 59.4) and (23.6, 5.8). The
point of most rapid decrease on the
graph of P is (14.2, 59.4). (The other
inflection point is a point of least
rapid decrease.) The most rapid
decrease occurs at 14.2°C, when
59.4% of eggs hatch. At this
temperature, P′(14.2) ≈ −11.1 , so the
percentage of eggs hatching is
declining by 11.1 percentage points
per °C. A small increase in
temperature will result in a relatively
large increase in the percentage of
eggs not hatching.
30. Home Sale
a. We solve the equation H ′′( x) = 0 and obtain the solution x ≈ 14.11 . A look at the graph
indicates that this point corresponds to the least rapid increase. The median house size was
increasing least rapidly in 1994 ( x ≈ 14.11 ). At the time, the median house size was
approximately H (14.11) ≈ 1938 square feet and was increasing at a rate of approximately
H ′(14.11) ≈ 7.3 square feet per year.
b.
Square
footage
2200
2100
2000
1900
1800
1700
7
9
11
13
15
17
19
21
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Section 4.3: Inflection Points 163
Calculus Concepts
80
H'(x)
60
40
20
0
7
9
11
13
15
17
19
21
20
H''(x)
10
0
-10
7
9
11
13
15
17
19
21
-20
c.
We compare the slopes of the graph of H at the endpoints corresponding to x = 7 and x =
21:
H ′(7) ≈ 61.7 square feet per year
H ′(21) ≈ 58.4 square feet per year
Thus the median house size was increasing most rapidly in 1987.
31. Price
a. The relative maximum point on the derivative graph between the values x = 4 and x = 10
occurs where x ≅ 5.785. This is the inflection point on the original function. The relative
minimum point on the derivative graph between the values x = 4 and x = 10 occurs where x
≅ 8.115. This is the other inflection point on the original function.
b. The relative max and min on the first derivative graph correspond to the x-intercepts on the
second derivative graph.
c. According to the model, between 1990 and 2001, the gas prices were declining most rapidly
in 1990, and they were increasing most rapidly in 2001. This is easiest to see if one examines
the first derivative graph.
d. According to the model, between 1994 and 2000, the gas price was decreasing most rapidly
in 1999 (approximately where x = 8.115) and it was increasing most rapidly in 1996
(approximately where x = 5.785).
32. Cable TV
P ( x) = 6 +
62.7
1 + 38.7e −0.258 x
Using the formula
P ( x) =
LABe− Bx
(1 + Ae− Bx ) 2
626.034e −0.258 x
(1 + 38.7e
percent x years after the end of 1970
)
− 0.258 x 2
for the derivative, we have
percentage points per year x years after the end of 1970
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164
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
The inflection point occurs where P′( x) is a maximum, at x ≈ 14.1. The percentage was
increasing most rapidly in the first half of 1985, when the percentage was P(14.1) ≈ 31.1% and
the rate of change of the percentage was P(14.1) ≈ 4.04 percentage points per year.
33. Donors
D (t ) = −10.247t 3 + 208.114t 2 − 168.805t + 9775.035 donors t years after 1975
D′(t ) = −30.741t 2 + 416.288t − 168.805 donors per year t years after 1975
a. Using technology, we find that (0.418, 9740.089) is the approximate relative minimum
point, and (13.121, 20,242.033) is the approximate relative maximum point on the cubic
model.
b. The inflection point occurs where D′(t ) has its maximum, at t ≈ 6.770. The inflection
point is approximately (6.8, 14,991.1).
c. i.
ii.
Because 6.8 is between t = 6 (the end of 1981) and t = 7 (the end of 1982), the
inflection point occurs during 1982, shortly after the team won the National
Championship. This is when the number of donors was increasing most rapidly.
The relative maximum occurred around the same time that a new coach was hired.
After this time, the number of donors declined.
34. Cable TV
a.
A( x ) = −0.126 x 3 + 1596
. x 2 + 1802
.
x + 40.930
annual dollars per person x years after 1984
A′( x) = −0.378 x 2 + 3.192 x + 1.802 annual
dollars per person per year x years after 1984
A′′( x) = −0.756 x + 3.192 annual dollars per
person per year per year x years after 1984
The inflection point on the graph of A is
approximately (4.222, 67.507). The
corresponding point on the graph of A′ is
the relative maximum, approximately (4.222,
8.541). The corresponding point on the graph
of A′′ is the x-intercept, approximately
(4.222, 0).
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Section 4.3: Inflection Points 165
Calculus Concepts
b. For integer values of x, the maximum value
of A′( x) is A′(4) ≈ 8.522 . In 1988, the rate
of change of the annual amount spent per
person was $8.52 per year.
35. Labor
a. N (h) =
62
1 + 11.49e−0.654 h
derivative, we have
N ′(h) =
components after h hours. Using the formula
(
62(−0.654)(11.49) e −0.654 h
(1 + 11.49e
)
−0.654 h 2
LABe− Bx
(1 + Ae− Bx ) 2
for the
) = 465.89652e−0.654h 1 + 11.49e−0.654h −2
(
)
components per hour after h hours.
The greatest rate occurs when N ′(h) is maximized, at h ≈ 3.733 hours, or approximately
3 hours and 44 minutes after she began working.
b. Her employer may wish to give her a break after 4 hours to prevent a decline in her
productivity.
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166
Chapter 4: Analyzing Change: Applications of Derivatives
36. Lake Level
a.
(
)
(
)
Calculus Concepts
L(d ) = −5.345 ⋅ 10−7 d 3 + 2.543 ⋅ 10−4 d 2 − 0.0192d + 6226.192 feet above sea level
(
)
(
)
d days after September 30, 1995; L′(d ) = −1.6035 ⋅ 10−6 d 2 + 5.086 ⋅ 10−4 d − 0.0192
feet per day d days after September 30, 1995; Because 1996 was a leap year, the number of
days from October 1, 1995, to July 31, 1996, was 31 + 30 + 31 + 31 + 29 + 31 + 30 + 31 +
30 + 31 = 305 days. The lake level was rising most rapidly when L′(d ) was a maximized at
159 days after the end of September 30, 1995 which occurred on March 7, 1996.
b. One possible answer: A spring thaw (maybe in conjunction with rain) probably caused the
lake level to rise rapidly.
c. One possible answer: Yes, because melting snow and rainfall are likely to follow a similar
pattern each year.
37. Labor
a.
b.
H ( w) =
10,111102
.
1 + 1153.222e − 0.727966w
(
total labor-hours after w weeks
H ′ ( w) = 10,111102
. ( −1) 1 + 1153.222e − 0.727966w
(
) (1153.22e
−2
≈ 8, 488,330.433e−0.727966 w 1 + 1153.222e−0.727966 w
)
−2
− 0.727966 w
)(−0.727966)
labor-hours per week
after w weeks
c.
The derivative gives the manager information
approximately the number of labor-hours spent
each week
d. The maximum point on the graph of H ′ is approximately (9.685, 1840.134). Keeping in
mind that the model must be discretely interpreted, we conclude that in the tenth week, the
most labor-hours are needed. That number is H ′(10) ≈ 1816 labor-hours.
e.
(
)(
)
 d − 0.727966w 
− 0.727966 w −2
H ′′( w) = 8,488,330.433
e
 1 + 1153.222e
 dx

(
) (
= 8,488,330.433(e
)(−0.727966)(1 + 1153.222e
+ 8,488,330.433(e
)(−2)(1 + 1153.222e
(1153.222e
) (–0.727966)
)
)
)
−2 
d
+ 8,488,330.433 e − 0.727966w 
1 + 1153.222e − 0.727966w

 dx

− 0.727966 w
− 0.727966 w
− 0.727966 w −2
− 0.727966 w −3
−0.727966 w
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.3: Inflection Points 167
Calculus Concepts
(
)(
)
−2
≈ −6,179,214.512 e − 0.727966w 1 + 1153.222e − 0.727966w
(
+ 1.4252 ⋅ 1010
) (e
−1.455932 w
)(1 + 1153.222e
)
− 0.72766 w −3
Use technology to find the maximum of H ′′( w) , which occurs at w ≈ 7.876. The point of
most rapid increase on the graph of H ′ is (7.876, 1226.756). This occurs approximately 8
weeks into the job, and the number of labor-hours per week is increasing by approximately
H ′′(8) ≈ 513 labor-hours per week per week.
f.
Use technology to find the minimum of the graph of H ′′ , which occurs at w ≈ 11.494. The
point of most rapid decrease on the graph of H ′ is (11.494, 1226.756). This occurs
approximately 12 weeks into the job when the number of labor-hours per week is changing
by approximately H ′′(12) = −486 labor-hours per week per week.
g. By solving the equation H ′′′( w) = 0 , we can find the input values that correspond to a
maximum or minimum point on the graph of H ′′ , which corresponds to inflection points on
the graph of H ′ , the weekly labor-hour curve.
h. Since the minimum of H ′′( w) occurs approximately 4 weeks after the maximum of H ′′( w) ,
the second job should begin approximately 4 weeks into the first job.
38. Advertising
a.
P ( x) =
57454.128
1 + 31.876e −0.090864 x
b. Using the formula
P′( x) ≈
(
dollars where x is the number of labor hours
LABe− Bx
(1 + Ae− Bx ) 2
166410.3e−0.090864 x
1 + 31.876e −0.090864 x
)
2
for the derivative, we have
dollars per labor hour
where x is the number of labor hours. The inflection point occurs when P′( x) is a
maximum at x ≈ 38 labor hours. The profit is increasing most rapidly at 38 labor hours, the
profit is P(38) ≈ 28,727 dollars, and the rate of change is P′(38) ≈ 1305 dollars per labor
hour.
c. Answers may vary.
39. Refuse
a. Between 1980 and 1985, the average rate of change was smallest at
122 − 117
= 1 million tons per year.
1985 − 1980
b. g (t ) = 0.008t 3 − 0.347t 2 + 6.108t + 79.690 million tons t years after 1970
c.
g ′(t ) = 0.025t 2 − 0.693t + 6.108 million tons per year t years after 1970
g ′′(t ) = 0.051t − 0.693 million tons per year per year t years after 1970
Copyright © Houghton Mifflin Company. All rights reserved.
168
Chapter 4: Analyzing Change: Applications of Derivatives
d.
Calculus Concepts
g ′′(t ) = 0
0.0507t − 0.693 = 0
0.0507t = 0.693
t ≈ 13.684
Solving g ′′(t ) = 0 gives t ≈ 13.684, which corresponds to mid-1984. The corresponding
amount of garbage is g(13.684) ≈ 120 million tons and the corresponding rate of increase is
g ′(13.684) ≈ 1.4 million tons per year.
e.
Because the graph of g ′′ crosses
the t-axis at 13.68, we know that
input corresponds to an inflection
point of the graph of g. Because
the graph of g ′ has a minimum at
that same value, we know that it
corresponds to a point of slowest
increase on the graph of g.
f.
The year with the smallest rate of change is 1984, with g(14) ≈ 120.4 million tons of
. million tons per year.
garbage, increasing at a rate of g ′(14) ≈ 137
40. Revenue
a. Using symmetric difference quotients to estimate rates of change, we see that revenue was
growing most rapidly in 1998, when the rate of change was approximately
1713.1 − 1180.4
= 266.35 million dollars per year.
1999 − 1997
b.
R (t ) = −5.285t 3 + 121.667t 2 − 689.840t + 1912.979 million dollars of revenue
t years after 1990
c.
R ' (t ) = −15.854t 2 + 243.334t − 689.840 million dollars per year t years after 1990
R ' ' (t ) = −31.708t + 243.334 million dollars per year per year t years after 1990
d. Solving R′′(t ) = 0 gives t ≈ 7.7. For integer values of t, R′(t ) has its maximum at t = 18, in
2008. According to the model, the revenue were R(18) ≈ 1475.24 million, and the rate of
change was R’(18) ≈ 242.19 million dollars per year.
41. Reaction
a. The first differences are greatest between 6 and 10 minutes, indicating the most rapid
increase in activity.
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Section 4.3: Inflection Points 169
Calculus Concepts
b.
1.930
A(m) =
1 + 31720
. e − 0.439118m
U / 100 µL m minutes after the mixture reaches 95°C
The inflection point, whose input is found using technology to locate the maximum point
on a graph of A′ , is approximately (7.872, 0.965). After approximately 7.9 minutes, the
activity was approximately 0.97 U/100 µL and was increasing most rapidly at a rate of
approximately 0.212 U/100 µL/min.
42. Emissions
a.
(
) (
) (
)
) (
) (
)
N (t ) = −3.611 ⋅ 10−4 t 3 + 2.283 ⋅ 10−2 t 2 + 1.349 ⋅ 10−2 t + 6.990 millions of metric tons
t years after 1940
b.
(
N ′(t ) = −1.083 ⋅ 10−3 t 2 + 4.567 ⋅ 10−2 t + 1.349 ⋅ 10−2 millions of metric tons per year
t years after 1940
c. Using technology, we find that N ′(t ) is maximized at t ≈ 21.1. For integer values of t, the
maximum value of N ′(t ) is obtained at t = 21, which corresponds to the year 1961. The
amount of emissions was N(21) ≈ 14.0 million metric tons, and emissions were increasing
at the rate of N ′(21) ≈ 0.5 million metric tons per year.
43. The graph of f is always concave up. A parabola that opens upward fits this description.
44. Possible graphs are parabolas or lines.
45. a. The graph is concave up between x = 0 and x = 2, has an inflection point at x = 2 and is
concave down between x = 2 and x = 4.
b.
46. a. The graph is concave up between x = 0 and x = 2, has inflection points at x = 0 and x = 2
and is concave down to the left of x = 0 and to the right of x = 2.
b.
One possible graph is shown.
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170
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
47.
One possible answer: Cubic and logistic models have inflection points, as do some product,
quotient, and composite functions. Note: In the answer in the text exponential is listed
twice and cubic is omitted.
48.
One possible answer: Approximate values of inflection points can often be determined by
viewing a graph of the function. Accurate values may be determined by finding the input values
corresponding to a relative minimum or maximum of the function’s derivative. The second
derivative is always undefined or zero at an inflection point.
Section 4.4 Interconnected Change: Related Rates
1.
df
dx
=3
dt
dt
2. 2 p
3.
4.
dp
ds
=5
dx
dx
dk
dx
= 12 x
dy
dy
dy
dx
dx
dx
= 27 x 2
+ 24 x + 4
dt
dt
dt
dt
dy
dx
= 27 x 2 + 24 x + 4
dt
dt
(
)
5.
dg
dx
= 3e3 x
dt
dt
6.
2 dx
dg
= 30 xe15 x
dt
dt
7.
df
dx
= 62(ln1.02)(1.02 x )
dt
dt
8.
dp
5 ds
=
dx 7 + s dx
9.
dh
da
da
= 6 + 6ln a
dy
dy
dy
da
= 6(1 + ln a )
dy
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Calculus Concepts
Section 4.4: Interconnected Change: Related Rates
171
10. Write v as v = πh( xw + w2 )
dv
dw 
 dw
= πh  x
+ 2w

dt
dt 
 dt
dv
dw
= πh( x + 2 w)
dt
dt
11.
(
ds
= πr 12 r 2 + h 2
dt
πrh dh
=
r 2 + h 2 dt
−1
2
)
(2h)
dh
dt
dr 
 dh
12. 0 = 13 π  r 2
+ h 2r 
dt 
 dt
dh
dr
0 = r2
+ h 2r
dt
dt
dh
dr
r2
= −h 2r
dt
dt
dh −2h dr
=
dt
r dt
13. Use the Product Rule with πr as the first term and r 2 + h 2 as the second term.
−1
dr
dh   dr  2
2
2
0 = πr 12 r 2 + h 2
 2r + 2 h  +  π  r + h
dt   dt 
 dt
πr
dh 
dr
 dr
0=
r + h  + π r 2 + h2

dt 
dt
r 2 + h 2  dt
(
)
14. We apply the Product Rule twice:
dw
0 = πhw
+ (derivative of πhw)( x + w)
dt
dw
dw
dh
0 = πhw
+ ( πh
+ πw )( x + w)
dt
dt
dt
w = 31.54 + 12.97 ln 5 ≈ 52.4 gallons per day
dw 12.97 dg 12.97  2

b.
=
=
inches per year  ≈ 0.43 gallon per day per year

dt
g dt
5  12

The amount of water transpired is increasing by approximately 0.43 gallon per day per
year. In other words, in one year, the tree will be transpiring approximately 0.4 gallon more
each day than it currently is transpiring.
15. a.
16. a. 5 feet 8 inches = 68 inches = h
0.45w
0.45w
B=
=
for a 5′8′′ woman weighing w pounds
2
0.00064516(68 ) 2.98321984
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172
Chapter 4: Analyzing Change: Applications of Derivatives
b. For the specific equation in part a, we have
Calculus Concepts
dB
0.45
dw
=
.
dt 2.98321984 dt
c. Because the variable w does not appear in the equation in part b, the fact that the woman
weighs 160 pounds is not relevant information. We have dw = 1 pound per month, so
dt
dB
0.45
=
(1) ≈ 0.15 point per month.
dt 2.98321984
d. Because the variable B does not appear in the equation in part b, the fact that the woman
has body mass index of 24 points is not relevant information. We have dB = −0.1 point per
dt
0.45
dw
month, so −0.1 =
2.98321984 dt
dw (−0.1)2.98321984
=
≈ −0.7 pound per month
dt
0.45
17. a.
b.
B=
0.45(100)
0.00064516h
2
(
=
45
0.00064516h 2
points
)
dB
45
dh
45
dh
=
−2h −3
=
3
dt 0.00064516
dt 0.00064516h dt
c. Evaluate the equation in part b at h = 63 inches and dh = 0.5 inch per year to obtain
dt
dB ≈ −0.2789 point per year.
dt
18. a. We are given h = 53% (remains constant), t = 80 F and dt = 2 F per hour where x is time
dx
dA
measured in hours. The question asks for A and
. We find A as
dx
A = 2.70 + 0.885(80) − 78.7(0.53) + 1.20(80)(0.53) ≈ 83.7 F . Next, we treat h as a constant
and find the derivative of the apparent temperature function with respect to x:
dA = 0.885 dt + 1.2h dt
dx
dx
dx
After substituting the known values, we are able to solve for dA :
dx
dA = 0.885(2) + 1.2(0.53)(2) = 3.042 F per hour
dx
The apparent temperature is approximately 83.7 F and is increasing at a rate of
approximately 3.04 F per hour.
b. We are given t = 100 F (remains constant), h = 0.3 and dh = −0.02 per hour where x is
dx
time measured in hours. (Note that we must convert both the humidity and the rate of
change of humidity to decimals.) The question asks for A and dA . We find A as
dx
A = 2.70 + 0.885(100) − 78.7(0.30) + 1.20(100)(0.30) ≈ 103.6 F . Next, we treat t as a
constant and find the derivative of the apparent temperature function with respect to x:
dA = −78.7 dh + 1.2t dh
dx
dx
dx
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Calculus Concepts
Section 4.4: Interconnected Change: Related Rates
173
After substituting the known values, we are able to solve for dA :
dx
dA = −78.7(−0.02) + 1.2(100)(−0.02) = −0.826 F per hour
dx
The apparent temperature is approximately 103.6 F and is decreasing at a rate of
approximately 0.83 F per hour.
19. a. We know h = 32 feet, d = 10
foot, dh = 0.5 foot per year, and we wish to find dV . We
12
dt
dt
treat d as a constant and find the derivative with respect to time t to obtain the related rates
equation dV = 0.002198d 1.7399251.133187 h0.133187 dh .
dt
dt
dV
Substituting the values given above results in
≈ 0.0014 cubic foot per year.
dt
2 foot per year, and we wish to find dV . We treat
b. We know h = 34 feet, d = 1 foot, dd = 12
dt
dt
h as a constant and find the derivative with respect to time t to obtain the related rates
equation dV = 0.002198(1.739925d 0.739925 ) h1.133187 dd .
dt
dt
dV
Substituting the values given above results in
≈ 0.0347 cubic foot per year.
dt
20. a.
b.
11.56 P
megajoules per person, where P is the number of kilograms of wheat produced
K
per hectare per year and K is the carrying capacity of the crop in people per hectare
D=
dD
 −1 dK  11.56 dP
= 11.56 P  2
+
dt
K dt
 K dt 
c. In order to answer the question posed, we need to solve for dP in the equation
dt
 −1 dK  11.56 dP
2 = 11.56(10)  2
. Doing so requires that we know values of K and dK ,
+
dt
dt
K
dt
K

which we are not given. We interpret dP as the rate of change of the crop production with
dt
respect to time. This rate of change tells us how rapidly the yearly amount of wheat grown
is changing as time changes.
21. a.
b.
M


L=
0.4 
 48.10352 K 
dL 
M

=

dt  48.10352 
5
3
(
5
3
M


=

48.10352


5
3
K
−2
3
)
−2 K −53 dK
3
dt
c. We are given K = 47 and dK = 0.5. Using the fact that L = 8 and the original equation, we
dL
can find the value of M corresponding to the current levels of labor and
capital: M ≈ 781.39 . Substituting the known values into the equation in part b gives
dL ≈ −0.057 thousand worker-hours per year.
dt
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174
Chapter 4: Analyzing Change: Applications of Derivatives
22.
Calculus Concepts
We are given dx = 3 feet per second. We need to find dh when
dt
dt
h = 6 feet. Using the Pythagorean Theorem, we have the equation
x 2 + h 2 = 152 . Taking the derivative of this equation with respect to
time t in seconds, we have 2 x dx + 2h dh = 0 . The only variable we
dt
dt
lack is the value of x when h = 6 feet, which we find as
x 2 + 62 = 152
x 2 = 189
x = 189 ≈ 13.7 feet
− 189
Thus 2 189(3) + 2(6) dh = 0 or dh =
≈ −6.9 feet per second . The ladder is sliding
dt
dt
2
down the wall at a rate of approximately 7 feet per second. As the ladder gets close to the
ground, h approaches zero. Rewriting 2 x dx + 2h dh = 0 , we have dh = − x dx . As h
dt
dt
dt
h dt
approaches zero, the fractional term with h in the denominator approaches infinity. (Note that
the model does not take into account the resistance created as the ladder slides down the wall,
accounting for this unreasonable answer.) We cannot answer the question posed.
23.
We are told that dV = 2 feet per second and
dt
v = (500 yards)(3 feet per yard) = 1500 feet, and
we need to know dd . Converting 100 yards to
dt
feet and using the Pythagorean Theorem, we
know that v 2 + 3002 = d 2 . Taking the
derivatives of both sides with respect to time
gives 2v dv + 0 = 2d dd .
dt
dt
To solve for dd , we need to know the value of d when v = 1500 feet. Use the Pythagorean
dt
Theorem: 15002 + 3002 = d 2 to find that d ≈ 1529.71 feet. Thus we have
2v dv = 2d dd
dt
dt
9(1500)(2) = 2(1529.71) dd
dt
dd ≈ 1.96 feet per second
dt
The balloon is approximately 1529.7 feet from the observer, and that distance is increasing by
approximately 1.96 feet per second.
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Calculus Concepts
Section 4.4: Interconnected Change: Related Rates
24.
175
We are given s = 100 feet and ds = 2 feet per
dt
second where t is the time in seconds. We are
asked to find dh . We use the Pythagorean
dt
Theorem to obtain the equation h 2 + 802 = s 2 .
Differentiating with respect to t gives
2h dh = 2 s ds .
dt
dt
The only piece of information we lack is the value of h when s = 100 feet. We find this value
using the Pythagorean Theorem: h 2 + 802 = 1002 . Solving for h yields h = 60 feet. Thus we
have 2(60 feet) dh = 2(100 feet)(2 feet per second) or dh = 20
≈ 3.33 feet per second.
6
dt
dt
We are told that dd = 22 feet per second and
25.
dt
d = 30 feet. We wish to find dh . We use the
dt
Pythagorean Theorem: h = 602 + (60 − d ) 2
to obtain the related rates equation:
2
2h dh = 0 + 2(60 − d )( −1) dd
dt
dt
To find the value of h, we substitute d = 30 into
the Pythagorean Theorem: h 2 = 602 + 302
h ≈ 67.08
Thus we have
2h dh = −2(60 − d ) dd
dt
dt
dh
2(67.08) = −2(60 − 30)(22)
dt
dh ≈ 1320 ≈ 9.8 feet per second
dt 2(67.08)
The runner is approximately 67.1 feet from home plate, and that distance is decreasing by
approximately 9.84 feet per second.
26. a. The volume of a sphere with radius r is given by the equation V = 43 πr 3 . A diameter of 20
inches corresponds to a radius of 10 inches, so V = 43 π (103 ) ≈ 4188.79 cubic inches.
b. We are given dV = 5 cubic feet per minute where t is the time in minutes. Because the
dt
radius is given in inches, we will convert it to feet in order for the units to match:
r = 10
= 5 foot. Differentiating the volume equation gives dV = 43 (3πr 2 ) dr = 4 πr 2 dr .
12 6
dt
dt
dt
Substituting the known information into the related rates equation, we find that
2
2
5 = 4 π  56  dr or dr = 45π  65  ≈ 0.57 foot per minute.
dt
dt
 
 
27. a. The volume of a sphere with radius r centimeters is given by the formula V = 43 πr 3 cubic
centimeters. When r = 10, V ≈ 4188.79 cm3 .
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176
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
b. Differentiating the volume equation with respect to time t yields dV = 43 (3πr 2 ) dr .
dt
dt
dr
We find
as the average rate of change between the points (0, 12) and (30, 8):
dt
dr 8 − 12 −4
=
=
cm per minute
dt 30 − 0 30
Substituting r = 10 and dr = −304 into the related rates equation, we find that
dt
dV = 4 (3π(102 )) −4 ≈ −167.6 cm3 per minute
3
30
dt
28.
We are told that r = h = 2 inches and dh = 0.2 inch per day
dt
where t is the time the salt has been leaking in days. We are
asked to find dV and V. We assume that we can use the
dt
formula for the volume of a right circular cylinder:
V = 13 πr 2 h . Because the radius and height are always equal,
we replace r with h and obtain: V = 13 πh3 .
Differentiating the volume equation with respect to t gives dV = πh 2 dh . Substituting the given
dt
dt
values into the volume equation and the related rate equation, we have
V = 13 π(23 ) ≈ 8.4 cubic inches and dV = π(22 )(0.2) ≈ 2.51 cubic inches per day
dt
The salt is leaking out at a rate of approximately 2.51 cubic inches per day. When the height of
the pile is 2 inches, the amount of salt in the pile is approximately 8.4 cubic inches.
29.
We are told
dV 1 T 1 T  1 cm3 
=
=

=
dt sec sec  0.06T 
1 cm3
0.06
per second
We wish to find dh . The volume of a cone with radius r
dt
πr 2 h 3
cm . Because
3
15 = h or r = h .
of similar triangles, we know that 2.5
r
6
and height h, both in centimeters is V =
Substituting this expression into the volume equation gives
volume in terms of height: V =
π
( h6 )
3
2
h
=
πh3
108
dV 3πh 2 dh
.
=
dt
108 dt
2
1 , 1 = 3π(6 ) dh which gives dh ≈ 4.34 cm per second
When h = 6 cm and dV = 0.06
dt
0.06
108 dt
dt
Differentiating both sides with respect to time t gives
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Calculus Concepts
Section 4.4: Interconnected Change: Related Rates
177
30. We are given the equation pv = c where c is a constant, v = 75 cubic inches, p = 30 pounds per
dp
= 2 pounds per square inch per minute where t is the time in minutes
dt
sincethe pressure began increasing. We are asked to find dv . Differentiating the equation with
dt
dp
dv
respect to time t (using the Product Rule) gives p + v = 0 . Substituting the known values
dt
dt
dv
dv
dv
we have: 30 + 75(2) = 0 or
= −5 cubic inches per minute. The volume
and solving for
dt
dt
dt
square inch, and
is compressing at a rate of 5 cubic inches per minute.
31. One possible answer:
Begin by solving for h: h =
V
πr
2
=
V −2
r
π
Differentiate with respect to t (V is a constant):
Substitute πr 2 h for V:
dh −2h dr
=
dt
r dt
dr
r dh
Rewrite:
=
dt −2h dt
(
(
)
dh V
dr
=
−2r −3
dt π
dt
)
dh πr 2 h
dr
=
−2r −3
dt
π
dt
Simplify:
32. One possible answer: We must use implicit differentiation. Instead of having a function F(x)
changing with respect to x, we now wish to find how F(x) changes with respect to an
independent variable t.
33. One possible answer: The first step, wherein the independent and dependent variables are
identified, is the most critical step in solving the problem correctly because if the problem is not
set up properly, the correct answer will not be found.
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178
Chapter 4: Analyzing Change: Applications of Derivatives
Calculus Concepts
Chapter 4 Concept Review
1.
a.
T has a relative maximum
point at (0.682, 143.098)
and a relative minimum
point at (3.160, 120.687). These
points can be determined by
finding the values of x between 0
and 6 at which the graph of T ′
crosses the x-axis. (There is also a
relative maximum to the right of
x = 6.)
b.
T has two inflection points:
(1.762, 132.939) and
(5.143, 149.067). These points
can be determined by finding the
values of x between 0 and 6 at
which the graph of T ′′ crosses the
x-axis. These are also the points at
which T ′ has a relative maximum
and relative minimum.
c.
d. To determine the absolute maximum and minimum, we compare the outputs of the relative
extrema with the outputs at the endpoints x = 0 and x = 6. The number of tourists was
greatest in 1994 at 166.8 thousand tourists. The number was least in 1991 at 120.9
thousand.
e. To determine the greatest and least slopes, we compare the slopes at the inflection points
with the slopes at the endpoints x = 0 and x = 6. The number of tourists was increasing
most rapidly in 1993 at a rate of 23.1 thousand tourists per year. The number of tourists
was decreasing most rapidly in 1990 at a rate of 13.3 thousand tourists per year.
2. a. (4.5 thousand people per year)
b. 202 +
1
2
( 14 year ) = 1.125 thousand people
(4.5) = 204.25 thousand people
3. Step 1: Output quantity to be minimized: cost
Input quantities: distances x and y
Step 2: See Figure 5.27 in the Chapter 5 Review Test.
Step 3: Cost = 27x + 143y dollars for distances of x feet and y feet.
Step 4: Convert the distances in miles in the figure to distances in feet using the fact the
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Calculus Concepts
Chapter 4: Concept Review
179
1 mile = 5280 feet: 3.2 miles = 16,896 feet and 1.6 miles = 8448 feet. Using the
Pythagorean Theorem, we know that 84482 + (16,890 − x) 2 = y 2 . Solving for positive
y yields y = 84482 + (16,896 − x)2 .
Substituting this expression for y into the equation in Step 3 gives
C(x) = 27x + 143 84482 + (16,896 − x)2 dollars where x is the distance the pipe is run
on the ground.
Step 5: Input interval: 0 < x < 16,890
Step 6: The derivative of the cost function is
−1
dC
= 27 + 143 12 [84482 + (16,896 − x) 2 ] 2 2(16,896 − x)(−1)
dx
143(16,896 − x)
= 27 −
84482 + (16,896 − x)2
Setting this equal to zero and solving for x between 0 and 16,890 gives
x = 15,271.7 feet. Dividing this answer by 5280 feet per mile, we have an optimal
distance of x ≈ 2.89 miles
Step 7: Substituting the value of x in feet into the cost equation gives a cost of
C(15,271.7) ≈ 1,642,527.
4.
The derivative graph lying above the
axis to the left of zero and below the axis
to the right of zero indicates that the
graph of h increases to the left of zero
and decreases to the right of zero. Thus a
relative maximum occurs at x = 0.
The derivative graph indicates a
maximum slope of h between x = a and
x = 0 and a minimum slope of between
x = 0 and x = b. These points of extreme
slope are inflection points on the graph
of h.
5. Treating w as a constant and differentiating the function S with respect to time yields
dS = 0.000013w(2v) dv . We have w = 4000 pounds, v = 60 mph, and dv = −5 mph per second .
dt
dt
dt
Substituting these values into the related rates equation gives
dS = 0.000013(4000)(2 ⋅ 60)(−5) = −31.2 feet per second.
dt
The length of the skid marks is decreasing by 31.2 feet per second.
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