chapter 4 analyzing change: applications of...

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Chapter 4 Analyzing Change: Applications of Derivatives

Section 4.1 Approximating Change

1. 32% − (4 percentage points per hour)( ) 213 3hour 30 %=

2. 300 mph + (200 mph per hour)( ) 23

112hour 316= mph

3. (3.5) (3) (3)(0.5) 17 4.6(0.5) 19.3f f f ′≈ + = + = 4. (7.25) (7) (7)(0.25)

4 ( 12.9)(0.25) 0.775

g g g′≈ += + − =

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 year16 years of age of age

≈ =

Annual premiumfor 70-year-old $8000≈

Annual premium

$500for 72-year-old $8000 + (2 years) $9000

year

≈ =

(Estimates will vary.)

142 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts

Copyright © Houghton Mifflin Company. All rights reserved.

10.

Slope of tangent line

10 years2.5

4 decades≈ = years per decade

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 dollarsbillion per dollarsbillion 97dollarsbillion 1

dollarsbillion 97 =

(revenue dollars per sales dollars). Revenue is approximately $614 billion when $6 billion is spent on advertising. Revenue is approximately 5.662)97(5.0614 =+ 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.

Calculus Concepts Section 4.1: Approximating Change 143


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. )013087.1(79.268)( xxP = thousand people in year x

)013087.1)(013087.1(ln79.268)(' xxP = 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 12

(46.656)≈ or $23,328.

d. The percentage rate of change after 10 years is '( ) '(10)( ) (10)100% 100%

A t AA t A=i i = 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

bi100%= 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.



16. a. 3 2( ) 0.044 0.92 38G t t t= − + + points after t hours of study

(11) 90.8G ≈ points

2( ) 0.132 1.84G t t t′ = − + i. (11) 4.27G′ ≈ points per hour of study

ii. G(12) ≈ 94.4 points

b. (12) (11 1) (11) (11)(1)G G G G′= + ≈ + ≈ (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 = 3 2( ) 0.16 8.7 172 69.4C p p p p= − + + dollars when p units are produced hourly.

b. 2( ) 0.48 17.4 172C p p p′ = + + dollars per unit when p units are produced hour.

(5) 97C′ = dollars per unit (20) 16C′ = dollars per unit

(30) 82C′ = 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: (6) (5) $822.76 $731.90 $90.86C C− ≈ − = Cost of 21st unit: (21) (20) $1326.46 $1309.40 $17.06C C− ≈ − ≈

Cost of 31st unit: (31) (30) $1807.26 $1719.40 $87.86C C− ≈ − =

d. The model is concave down at x = 5, but it is concave up at x = 20 and x = 30.

d. 2 1( )( ) 0.16 8.7 172 69.4

C pA p p p p p

p−= = − + + dollars per unit

when p units are produced hourly



e. 1 2( ) 0.32 8.7 172 69.4A p p p p− −′ = − + + 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 (5) 30.08A′ ≈ dollars per unit per unit produced hourly. When the production level is at 20, the average cost is changing at a rate of (20) 6.47A′ ≈ − dollars per unit per unit produced hourly. When the production level is at 30, the average cost is changing at a rate of

(30) 6.71A′ ≈ 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. (12 ).06512

300(1 ) tA = +

b. 300(1.06697)tA = c. A(2) = $341.53 d. A’(2) = 300(ln 1.06697)(1.06697)2 = 22.14 dollars/year. e. A(2.25) (2) .25 '(2)A A≅ + = $347.07

20. a. (12 ).03212

2000(1 ) tA = +

b. 2000(1.0325)tA = c. A(5) = $2346.52 d. A’(5) = 2000(ln 1.0325)(1.0325)5 = 74.99 dollars/year. e. A(5.5) (5) .5 '(5)A A≅ + = $2384.02 21. Sales

a. ( )4 2( ) 7.032 10 1.666 47.130R x x x−= − ⋅ + + dollars when x hot dogs are sold, 100 < x < 1500.

b. Cost: xxC 5.0)( = dollars when x hot dogs are sold, 100 < x < 1500.

Profit: 130.47166.1)10032.7()()()( 24 ++⋅−=−= − xxxCxRxP dollars when x hot dogs are sold, 100 < x < 1500.

c. Marginal Revenue = R’(x) = -.0014x + 1.666 dollars/hot dog.

x (hot dogs)

′( )R x

(dollars per hot dog)

′( )c x

(dollars per hot dog)

′( )p x

(dollars per hot dog) 200 1.38 0.50 0.88 800 0.54 0.50 0.04 1100 0.12 0.50 −0.38 1400 −0.30 0.50 −0.80



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.

d.

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.

22. Production

a. 3 2( ) 0.068 2.933 55.269 146.983C x x x x= − + + dollars when x hundred balls are produced hourly

b. 2( ) 0.204 5.865 55.269C x x x′ = − + dollars per hundred balls when x hundred balls are produced hourly

(10) 16.99C′ ≈ 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. (3) 39.51C′ ≈ dollars per hundred balls or $0.395 per ball

(21) $21.94C′ ≈ 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.

d. 2 1( )( ) 0.068 2.933 55.269 149.983

C xA x x x x

x−= = − + + dollars per hundred balls

when x hundred balls are produced hourly

e. 2( ) 0.136 2.933 146.983A x x x−′ = − − dollars per hundred balls per hundred balls when x hundred balls are produced hourly

(3) 18.86A′ ≈ − dollars per hundred balls per additional hundred balls produced hourly or –$0.00189 per ball per additional ball produced hourly



(17) 1.13A′ ≈ − 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. 3 2( ) 0.109 1.555 10.927 100.320A t t t t= − + + t years after 1980

b. 2( ) 0.327 3.111 10.927A t t t′ = − + index points per year t years after 1980

(7) 5.2A′ ≈ index points per year

c. 1988 CPI estimate: (CPI in 1987) + (7)A′ (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. 3 2( ) 0.150 2.171 15.814 99.650C t t t t= − + + t years after 1980

b. 2( ) 0.450 4.343 15.814C t t t′ = − + index points per year t years after 1980

(7) 7.5C′ ≈ index points per year

c. 1988 CPI estimate: (CPI in 1987) + (7)C′ (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. ( ) 85.112(2.013252 )tP t = t years after 1980

b. ( ) 85.112(ln 2.013252)(2.013252 )

59.558(2.013252 ) index points per year years after 1980

t

t

P t

t

′ =

≈

(7) 7984P′ ≈ index points per year

c. 1988 CPI estimate: (CPI in 1987) + (7)P′ (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.



Brazil: a. ( ) 73,430(2.615939 )tB t = t years after 1980

b. ( ) 73.430(ln 2.615939)(2.615939 )

70.612(2.615939 ) index points per year years after 1980

t

t

B t

t

′ =

≈

(7) 59,193B′ ≈ index points per year

c. 1988 CPI estimate: (CPI in 1987) + (7)B′ (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 = 2( ) 12.16 254.28 105.60R x x x= − + − dollars, where x = the price for a large one-topping pizza, in dollars. 9.25 < x < 14.25. b. '( ) 24.32 254.28R x x= − + 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. 3 2( ) 0.158 5.235 23.056 154.884R A A A A= − + − + thousand dollars of revenue when A thousand dollars is spent on advertising. 5 < A < 19.

b. 3 2( ) 0.473 10.471 23.056R A A A′ = − + − thousand dollars of revenue per thousand dollars of advertising when A thousand dollars is spent on advertising

(10) 34.3R′ ≈ 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. (18) 12.0R′ ≈ 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.



26. Newspapers

a. 588.51457.3321.0008.0)( 23 ++−= xxxxn 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. 457.3642.0024.0)(' 2 +−= xxxn 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 0

( ) ( )'( ) lim

h

f x h f xf x

h→

+ −= . Assuming h is relatively close

to zero, ( ) ( )

'( )f x h f x

f xh

+ −≈ . Multiplying both sides of this approximation by h yields

'( ) ( ) ( )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.



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.

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.

9. One possible answer: One such graph is y x= 3 , 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, ( ) 0,f x′ < 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.



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.



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 + 1t

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 49F° .

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



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 7 2 4( ) 3( 5.345 10 ) 2(2.543 10 ) 0.0192 0L d d d− −′ = − ⋅ + ⋅ − = , and we obtain the solutions

43.8, 273.4d ≈ . 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. 2( ) 0.181 8.463 147.376S x x x= − + 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 ( ) 316.765(0.949 )pR p = dozen roses when the price per dozen is p dollars.

b. Multiply R(p) by the price, p. The consumer expenditure is ( ) 316.765 (0.949 )pE p 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 ( ) ( ) 6 ( )F p E p R p= − 316.765( 6)(0.949 ).pp= − 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.



32. Demand

a. A quadratic model for the data is 2( ) 0.0866 10.125 299.710p x x x= − + 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.

3 2( ) 0.0866 10.125 299.710R x x x 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. 3 2( ) 0.008 0.347 6.108 79.690G t t t t= − + + million tons of garbage taken to a landfill t years after 1975

b. 2( ) 0.025 0.693 6.108G t t t′ = − + million tons of garbage per year t years after 1975

c. In 2005 the amount of garbage was increasing by (30) 8.1G′ ≈ million tons per year.

d.

Because the derivative graph exists for all input values and never crosses the horizontal axis, G(t) has no relative maxima.

34. Price

a. Exponential model: ( ) 568.074(0.965582 )pA p = tickets sold on average when the price is p dollars

Quadratic model: 2( ) 0.15 16.007 543.286A p p p= − + 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.

( ) 568.074 (0.965582 )pR p 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 (28.55) 209A ≈ .



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

2

2 2 2

2 (2 3) 4 10

( 2) 2

x x x xy

x x

− − + −′ = + =+ +

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.

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 corresponding 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

minimum is [ ]y = − + +2 3 1549 1549 35 15492 2( . ) ( . ) ( . . ) ≈ −6.312.

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.

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)



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. 3.9 3.9 2'( ) 131.04 (1 2.1 )t tL t e e −= − + ; L”(t) = ( ) ( )

7.8 3.9

3 23.9 3.9

2146.4352 511.056

1 2.1 1 2.1"( )

t t

t t

e e

e eL t

−+

+ +=

18. 0.02 0.02 2'( ) 199.2 (1 99.6 ) t tL t e e− − −= + ; ( ) ( )

.04 .02

3 2.02 .02

7.936 3.984

1 99.6 1 99.6"( )

t t

t t

e e

e eL t

− −

− −

−+

+ +=

19. 2'( ) 3 12 2f x x x= − + ; "( ) 6 12 0 f x x= − = at x = 2

20. 2'( ) -0.3 2.4 3.6g t t t= + + ; "( ) -0.6 2.4 0g t t= + = 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 3 20.04 0.88 4.81 12.11x x x= − + +

2( ) 0.12 1.76 4.81g x x x′ = − +



b. ′′ = −g x x( ) . .0 24 176

The inflection point on the graph of g is approximately (7.333, 15.834). This is a point of most rapid decline.

26. a. f xe x

( ).

=+ −

20

1 19 0 5 . Using the formula 2(1 )

Bx

Bx

LABe

Ae

−

−+ for the derivative, we have

( )( )

( )0.5

20.5 0.520.5

20(1)(0.5) 19( ) 190 1 19

1 19

xx x

x

ef x e e

e

−−− −

−′ = = +

+

( ) ( ) ( )′′ =

+ + +

− − − − − −f x

d

dxe e e

d

dxex x x x( ) . . . .190 1 19 190 1 190 5 0 5 2 0 5 0 5 2

( ) ( ) ( ) ( )2 30.5 0.5 0.5 0.5 0.5190 ( 0.5) 1 19 ( 2) 1 19 19 ( 0.5)x x x x xe e e e e− −− − − − − = − + + − + −

( ) ( )= − + + +− − − − − −95 1 19 3610 1 190 5 0 5 2 0 5 3

e e e ex x x x. . .



b. The inflection point of f is approximately (5.889, 10). This is a point of most rapid increase.

27. Study Time

a. P te t

( ). .

=+ −

45

1 594 0 969125 ( )= + − −45 1 594 0 969125 1

. .e t percent after studying for t hours

( ) ( )′ = − + −− − −P t e et t( ) ( ) . . ( . ). .45 1 1 594 594 0 9691250 969125 2 0 969125

( ) 20.969125 0.969125259.0471125 1 5.94 percentage points per hour

after studying for hours

t te et

−− −= +

( ) ( )′′ =

+− − −

P td

dxe et t( ) . .. .259 0471125 1 5940 969125 0 969125 2

+ −2590471125 0 969125. .e t ( )d

dxe t1 594 0 969125 2

+

− −. .

( ) ( )= − +− − −259 0471125 0 969125 1 5940 969125 0 969125 2

. ( . ) .. .e et t

+ −259 0471125 0 969125. .e t ( ) ( )( ) . . ( . ). .− + −− − −2 1 594 594 0 9691250 969125 3 0 969125e et t



( )≈ − +− − −251049033 1 5940 969125 0 969125 2

. .. .e et t

+ +2982 462511 1 59493825 969125. ( . )–1. –0. –3e et t percentage points per hour per hour after studying for t hours

Solving ′′ =P t( ) 0 for t gives t ≈ 1838. . The inflection point on P is approximately (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 x x x( ) ( . ) ( . ) ( . ) .4 1619 10 3 1675 10 2 0 050 0 3085 3 3 2 percentage points per year x years after the end of 2000

5 2 3( ) 12(1.619 10 ) 6(1.675 10 ) 2(0.050)p x x x− −′′ = ⋅ − ⋅ + percentage points per year per year

x years after the end of 2000 Solving for x in the equation ( ) 0p x′′ = gives 13.44, 38.29x ≈ . 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 13.44x ≈ with the slope values at the endpoints: (0) 0.308p′ = − and (50) 0.225p′ ≈ . Both of these values are less

than the slope at 13.44x ≈ . 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, 38.29x ≈ , 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 38.29x ≈ . 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. 4 3 2( ) 0.00645 0.488 12.991 136.560 395.154P t t t t t= − + − + − percent when the temperature is t°C

3 2( ) 0.0258 1.464 25.982 136.560P t t t t′ = − + − + percentage points per °C when the temperature is t°C

2( ) 0.0774 2.928 25.982P t t t′′ = − + − percentage points per °C per °C when the temperature is t°C



b. Because the graph ofP′′ 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, (14.2) 11.1P′ ≈ − , 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 ( ) 0H x′′ = and obtain the solution 11.14≈x . 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 ( 11.14≈x ). At the time, the median house size was approximately ≈)11.14(H 1938 square feet and was increasing at a rate of approximately

≈′ )11.14(H 7.3 square feet per year.

b.

1700 1800 1900 2000 2100 2200

7 9 11 13 15 17 19 21

Square footage



0

20

40

60

80

7 9 11 13 15 17 19 21H

'(x)

-20

-10

0

10

20

7 9 11 13 15 17 19 21H''(

x)

c. We compare the slopes of the graph of H at the endpoints corresponding to x = 7 and x =

21: 7.61)7( ≈′H square feet per year

≈′ )21(H 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

xe

xP258.07.381

7.626)( −+

+= percent x years after the end of 1970

Using the formula 2(1 )

Bx

Bx

LABe

Ae

−


( )2258.0

258.0

7.381

034.626)(

x

x

e

exP

−

−

+= percentage points per year x years after the end of 1970



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

3 2( ) 10.247 208.114 168.805 9775.035D t t t t= − + − + donors t years after 1975 2( ) 30.741 416.288 168.805D t t t′ = − + − 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. 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.

ii. 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 x x x( ) . . . .= − + + +0126 1596 1802 40 9303 2

annual dollars per person x years after 1984

2( ) 0.378 3.192 1.802A x x x′ = − + + annual dollars per person per year x years after 1984

( ) 0.756 3.192A x x′′ = − + 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).



b. For integer values of x, the maximum value of ( )A x′ is (4) 8.522A′ ≈ . In 1988, the rate of change of the annual amount spent per person was $8.52 per year.

35. Labor

a. 0.654

62( )

1 11.49 hN h

e−=+

components after h hours. Using the formula 2(1 )

Bx

Bx

LABe

Ae

−

−+ for the

derivative, we have

( )( )

( )0.654

20.654 0.65420.654

62( 0.654)(11.49)( ) 465.89652 1 11.49

1 11.49

hh h

h

eN h e e

e

−−− −

−

−′ = = +

+

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.



36. Lake Level

a. ( ) ( )7 3 4 2( ) 5.345 10 2.543 10 0.0192 6226.192L d d d d− −= − ⋅ + ⋅ − + feet above sea level

d days after September 30, 1995; ( ) ( )6 2 4( ) 1.6035 10 5.086 10 0.0192L d d d− −′ = − ⋅ + ⋅ −

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. H we w

( ), .

. .=

+ −10 111102

1 1153222 0 727966 total labor-hours after w weeks

b. ( ) ( )′ = − + −− − −H w e ew w( ) , . ( ) . . ( . ). .10 111102 1 1 1153222 115322 0 7279660 727966 2 0 727966

( ) 20.727966 0.7279668,488,330.433 1 1153.222 labor-hours per weekafter weeks

w we ew

−− −≈ +

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 (10) 1816H ′ ≈ labor-hours.

e. ( ) ( )′′ =

+− − −

H wd

dxe ew w( ) , , . .. .8 488 330 433 1 11532220 727966 0 727966 2

( ) ( )+ +

− − −8 488 330 433 1 11532220 727966 0 727966 2, , . .. .e

d

dxew w

( ) ( )= − +− − −8 488 330 433 0 727966 1 11532220 727966 0 727966 2, , . ( . ) .. .e ew w

( ) ( )+ − +− − −8 488 330 433 2 1 11532220 727966 0 727966 3, , . ( ) .. .e ew w

( )1153222 0 727966. .e w−(–0.727966)



( )( )≈ − +− − −6 179 214 512 1 11532220 727966 0 727966 2, , . .. .e ew w

( )+ ⋅14252 1010. ( )( )e ew w− − −+1455932 0 72766 3

1 1153222. ..

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

(8) 513H ′′ ≈ 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 (12) 486H ′′ = − labor-hours per week per week.

g. By solving the equation ( ) 0H w′′′ = , we can find the input values that correspond to a

maximum or minimum point on the graph ofH ′′ , 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. 0.090864

57454.128( )

1 31.876 xP x

e−=+

dollars where x is the number of labor hours

b. Using the formula 2(1 )

Bx

Bx

LABe

Ae

−


( )0.090864

20.090864

166410.3( )

1 31.876

x

x

eP x

e

−

−′ ≈

+ 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 (38) 1305P′ ≈ 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

1985 19801

−−

= million tons per year.

b. 3 2( ) 0.008 0.347 6.108 79.690g t t t t= − + + million tons t years after 1970

c. 2( ) 0.025 0.693 6.108g t t t′ = − + million tons per year t years after 1970

( ) 0.051 0.693g t t′′ = − million tons per year per year t years after 1970



d. ( ) 0

0.0507 0.693 0

0.0507 0.693

13.684

g t

t

t

t

′′ =− =

=≈

Solving ( ) 0g t′′ = 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

(13.684) 1.4g′ ≈ 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 garbage, increasing at a rate of ′ ≈g ( ) .14 137 million tons per year.

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

35.26619971999

4.11801.1713 =−−

million dollars per year.

b. 979.1912840.689667.121285.5)( 23 +−+−= ttttR million dollars of revenue

t years after 1990

c. 840.689334.243854.15)(' 2 −+−= tttR million dollars per year t years after 1990 334.243708.31)('' +−= ttR million dollars per year per year t years after 1990

d. Solving ( ) 0R t′′ = 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.



b. A me m

( ).

. .=

+ −1930

1 31720 0 439118 U / 100 Lµ 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. ( ) ( ) ( )4 3 2 2 2( ) 3.611 10 2.283 10 1.349 10 6.990N t t t t− − −= − ⋅ + ⋅ + ⋅ + millions of metric tons

t years after 1940

b. ( ) ( ) ( )3 2 2 2( ) 1.083 10 4.567 10 1.349 10N t t t− − −′ = − ⋅ + ⋅ + ⋅ 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 (21) 0.5N′ ≈ 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.



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. 3df dx

dt dt=

2. 2 5dp ds

pdx dx

=

3. 12dk dx

xdy dy

=

4. 227 24 4dy dx dx dx

x xdt dt dt dt

= + +

( )227 24 4dy dx

x xdt dt

= + +

5. 33 xdg dxe

dt dt=

6. 21530 xdg dx

xedt dt

=

7. 62(ln1.02)(1.02 )xdf dx

dt dt=

8. 5

7

dp ds

dx s dx=

+

9. 6 6ln

6(1 ln )

dh da daa

dy dy dy

daa

dy

= +

= +

Calculus Concepts Section 4.4: Interconnected Change: Related Rates 171


10. Write v as 2π ( )v h xw w= +

π 2dv dw dw

h x wdt dt dt

= +

π ( 2 )dv dw

h x wdt dt

= +

11. ( ) 122 21

2

2 2

π (2 )

π =

ds dhr r h h

dt dtrh dh

dtr h

−= +

+

12. 213

0 π 2dh dr

r h rdt dt

= +

20 2dh dr

r h rdt dt

= +

2 2dh dr

r h rdt dt

= −

2dh h dr

dt r dt

−=

13. Use the Product Rule with πr as the first term and 2 2r h+ as the second term.

( ) 122 2 2 21

20 π 2 2 π

dr dh drr r h r h r h

dt dt dt

− = + + + +

2 2

2 2

π0 π

r dr dh drr h r h

dt dt dtr h

= + + + +

14. We apply the Product Rule twice:

0 π (derivative of π )( )dw

hw hw x wdt

= + +

0 π (π π )( )dw dw dh

hw h w x wdt dt dt

= + + +

15. a. 31.54 12.97 ln5 52.4w = + ≈ gallons per day

b. 12.97 12.97 2

inches per year 0.43 gallon per day per year5 12

dw dg

dt g dt = = ≈

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.

16. a. 5 feet 8 inches = 68 inches = h

2

0.45 0.45

2.983219840.00064516(68 )

w wB = = for a 5 8′ ′′ woman weighing w pounds



b. For the specific equation in part a, we have0.45

2.98321984

dB dw

dt 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 dwdt

= 1 pound per month, so

0.45

(1) 0.152.98321984

dB

dt= ≈ point per month.

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 0.1dBdt

= − point per

month, so 0.45

0.12.98321984( 0.1)2.98321984

0.7 pound per month0.45

dw

dtdw

dt

− =

−= ≈ −

17. a. 2 2

0.45(100) 45 points

0.00064516 0.00064516B

h h= =

b. ( )33

45 452

0.00064516 0.00064516

dB dh dhh

dt dt dth−= − =

c. Evaluate the equation in part b at h = 63 inches and 0.5dhdt

= inch per year to obtain

0.2789dBdt

≈ − point per year.

18. a. We are given h = 53% (remains constant), 80 Ft = � and 2 F per hourdtdx

= � where x is time

measured in hours. The question asks for A and dAdx

. We find A as

2.70 0.885(80) 78.7(0.53) 1.20(80)(0.53) 83.7 FA = + − + ≈ � . Next, we treat h as a constant and find the derivative of the apparent temperature function with respect to x:

0.885 1.2dA dt dtdx dx dx

h= +

After substituting the known values, we are able to solve for dAdx

:

0.885(2) 1.2(0.53)(2) 3.042 F per hourdAdx

= + = �

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 100 Ft = � (remains constant), h = 0.3 and 0.02 per hourdhdx

= − where x is

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 dAdx

. We find A as

2.70 0.885(100) 78.7(0.30) 1.20(100)(0.30) 103.6 FA = + − + ≈ � . Next, we treat t as a constant and find the derivative of the apparent temperature function with respect to x:

78.7 1.2dA dh dhdx dx dx

t= − +



After substituting the known values, we are able to solve for dAdx

:

78.7( 0.02) 1.2(100)( 0.02) 0.826 F per hourdAdx

= − − + − = − �

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 = 1012 foot, 0.5dh

dt= foot per year, and we wish to find dV

dt. We

treat d as a constant and find the derivative with respect to time t to obtain the related rates

equation 1.739925 0.1331870.002198 1.133187dV dhdt dt

d h= .

Substituting the values given above results in 0.0014dVdt

≈ cubic foot per year.

b. We know h = 34 feet, d = 1 foot, 212

dddt

= foot per year, and we wish to find dVdt

. We treat

h as a constant and find the derivative with respect to time t to obtain the related rates

equation 0.739925 1.1331870.002198(1.739925 )dV dddt dt

d h= .

Substituting the values given above results in 0.0347dVdt

≈ cubic foot per year.

20. a. 11.56P

DK

= megajoules per person, where P is the number of kilograms of wheat produced

per hectare per year and K is the carrying capacity of the crop in people per hectare

b. 2

1 11.5611.56

dD dK dPP

dt dt K dtK

− = +

c. In order to answer the question posed, we need to solve for dPdt in the equation

2

1 11.562 11.56(10)

dK dP

dt K dtK

− = +

. Doing so requires that we know values of K and dKdt ,

which we are not given. We interpret dPdt as the rate of change of the crop production with

respect to time. This rate of change tells us how rapidly the yearly amount of wheat grown is changing as time changes.

21. a. 5 53 3 2

30.4 48.1035248.10352

M ML K

K

− = =

b. ( )53 5

32348.10352

dL M dKK

dt dt

−− =

c. We are given K = 47 and dKdL

= 0.5. Using the fact that L = 8 and the original equation, we

can find the value of M corresponding to the current levels of labor and capital: 781.39M ≈ . Substituting the known values into the equation in part b gives

0.057dLdt

≈ − thousand worker-hours per year.



22.

We are given 3dxdt

= feet per second. We need to find dhdt

when

h = 6 feet. Using the Pythagorean Theorem, we have the equation 2 2 215x h+ = . Taking the derivative of this equation with respect to

time t in seconds, we have2 2 0dx dhdt dt

x h+ = . The only variable we

lack is the value of x when h = 6 feet, which we find as

2 2 2

2

6 15

189

189 13.7 feet

x

x

x

+ =

=

= ≈

Thus 189

2 189(3) 2(6) 0 or 6.9 feet per second2

dh dhdt dt

−+ = = ≈ − . The ladder is sliding

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 2 0dx dhdt dt

x h+ = , we have dh x dxdt h dt

−= . As h

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 dVdt

= 2 feet per second and

v = (500 yards)(3 feet per yard) = 1500 feet, and

we need to know dddt

. Converting 100 yards to

feet and using the Pythagorean Theorem, we

know that 2 2 2300v d+ = . Taking the derivatives of both sides with respect to time

gives 2 0 2dv dddt dt

v d+ = .

To solve for dddt

, we need to know the value of d when v = 1500 feet. Use the Pythagorean

Theorem: 2 2 21500 300 d+ = to find that 1529.71d ≈ feet. Thus we have

2 2

9(1500)(2) 2(1529.71)

1.96 feet per second

dv dddt dt

dddt

dddt

v d=

=

≈

The balloon is approximately 1529.7 feet from the observer, and that distance is increasing by approximately 1.96 feet per second.



24.

We are given s = 100 feet and dsdt

= 2 feet per

second where t is the time in seconds. We are

asked to find dhdt

. We use the Pythagorean

Theorem to obtain the equation 2 2 280h s+ = . Differentiating with respect to t gives

2 2dh dsdt dt

h s= .

The only piece of information we lack is the value of h when s = 100 feet. We find this value

using the Pythagorean Theorem: 2 2 280 100h + = . Solving for h yields h = 60 feet. Thus we

have 2(60 feet) 2(100 feet)(2 feet per second)dhdt

= or 206 3.33dh

dt= ≈ feet per second.

25.

We are told that dddt

= 22 feet per second and

d = 30 feet. We wish to find dhdt

. We use the

Pythagorean Theorem: 2 2 260 (60 )h d= + − to obtain the related rates equation:

2 0 2(60 )( 1)dh dddt dt

h d= + − −

To find the value of h, we substitute d = 30 into

the Pythagorean Theorem: 2 2 260 30

67.08

h

h

= +≈

Thus we have 2 2(60 )

2(67.08) 2(60 30)(22)

13209.8 feet per second

2(67.08)

dh dddt dtdhdt

dhdt

h d= − −

= − −

≈ ≈

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 343πV r= . A diameter of 20

inches corresponds to a radius of 10 inches, so 343π(10 ) 4188.79V = ≈ cubic inches.

b. We are given 5dVdt = cubic feet per minute where t is the time in minutes. Because the

radius is given in inches, we will convert it to feet in order for the units to match: 10 512 6r = = foot. Differentiating the volume equation gives 2 24

3 (3π ) 4πdV dr drdt dt dtr r= = .

Substituting the known information into the related rates equation, we find that 25

65 4π drdt

=

or 25 6

54π 0.57drdt

= ≈

foot per minute.

27. a. The volume of a sphere with radius r centimeters is given by the formula 343 πV r= cubic

centimeters. When r = 10, V ≈ 4188.79 3cm .



b. Differentiating the volume equation with respect to time t yields 243

(3π )dV drdt dt

r= .

We find drdt

as the average rate of change between the points (0, 12) and (30, 8):

8 12 4

30 0 30

dr

dt

− −= =−

cm per minute

Substituting r = 10 and 430

drdt

−= into the related rates equation, we find that

2 343

430(3π(10 )) 167.6 cm per minutedV

dt−= ≈ −

28.

We are told that 2r h= = inches and 0.2dhdt

= inch per day

where t is the time the salt has been leaking in days. We are

asked to find dVdt

and V. We assume that we can use the

formula for the volume of a right circular cylinder: 21

3πV r h= . Because the radius and height are always equal,

we replace r with h and obtain: 313πV h= .

Differentiating the volume equation with respect to t gives 2π

dV dhdt dt

h= . Substituting the given

values into the volume equation and the related rate equation, we have

313π(2 ) 8.4V = ≈ cubic inches and 2

π(2 )(0.2) 2.51dVdt

= ≈ cubic inches per day

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 3

310.06

1 T 1 T 1 cmcm per second

sec sec 0.06T

dV

dt

= = =

We wish to find dhdt

. The volume of a cone with radius r

and height h, both in centimeters is 2

3πcm

3

r hV = . Because

of similar triangles, we know that 152.5

hr= or 6

hr = .

Substituting this expression into the volume equation gives

volume in terms of height: ( )2

36π π

3 108

h h hV = =

Differentiating both sides with respect to time t gives23π

108

dV h dh

dt dt= .

When h = 6 cm and 10.06

dVdt

= ,21 3π(6 )

0.06 108

dh

dt= which gives 4.34 cm per second

dh

dt≈



30. We are given the equation pv = c where c is a constant, v = 75 cubic inches, p = 30 pounds per

square inch, and 2dpdt

= pounds per square inch per minute where t is the time in minutes

sincethe pressure began increasing. We are asked to find dvdt

. Differentiating the equation with

respect to time t (using the Product Rule) gives 0dpdvdt dt

p v+ = . Substituting the known values

and solving for dvdt

we have: 30 75(2) 0dvdt

+ = or dvdt

= −5 cubic inches per minute. The volume

is compressing at a rate of 5 cubic inches per minute. 31. One possible answer:

Begin by solving for h: 22 ππ

V Vh r

r−= =

Differentiate with respect to t (V is a constant): ( )32π

dh V drr

dt dt−= −

Substitute 2πr h for V: ( )

23π

2π

dh r h drr

dt dt−= −

Simplify: 2dh h dr

dt r dt

−=

Rewrite: 2

dr r dh

dt h dt=

−

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.



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.)

c.

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.

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)( )14

year = 1.125 thousand people

b. 202 + 12 (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

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 2 2 28448 (16,890 )x y+ − = . Solving for positive

y yields 2 28448 (16,896 )y x= + − .

Substituting this expression for y into the equation in Step 3 gives

C(x) = 27x + 2 2143 8448 (16,896 )x+ − 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 122 21

2

2 2

27 143 [8448 (16,896 ) ] 2(16,896 )( 1)

143(16,896 )27

8448 (16,896 )

dCx x

dxx

x

−= + + − − −

−= −+ −

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

0.000013 (2 )dS dvdt dt

w v= . We have w = 4000 pounds, v = 60 mph, and 5 mph per seconddvdt

= − .

Substituting these values into the related rates equation gives

0.000013(4000)(2 60)( 5) 31.2dSdt

= ⋅ − = − feet per second.

The length of the skid marks is decreasing by 31.2 feet per second.

chapter 4 analyzing change: applications of...

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