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Chapter 6 Solutions

Worked solutions to all thirty problems, with the reason for each step. These explanations were added for this edition; the chapter answer key and original worked examples follow below.

Chapter problems → · Chapter answer key → · Original worked examples →

Problem 1

Original problem · clarified · added worked solution

You open up a business using $10 000 of your hard-earned money. You think that the rate of return on your investment (as a decimal) can be modeled by the equation

r = e^{- t} + 0.07

. What is the rate of increase of the assets in your business after 3 years? Use units of dollars per year.

Clarification: Use the asset model $A (t) = 10000 e^{t (e^{- t} + 0.07)}$ . Here the given r(t) is inserted into the exponent t r(t); it is not an instantaneous proportional growth rate.

Authored worked explanation for the original source problem.

Source review note: The source gives −42.40 dollars/year. The explicitly stated inserted-rate model gives about −423.62; a literal instantaneous-return interpretation gives positive growth instead. The original problem needs a model clarification.

If we use the source's constant-rate balance form with the varying rate inserted, the exponent is $h (t) = t (e^{- t} + 0.07)$ .
Why: The wording does not fully specify the asset model, so this assumption must be explicit.
$h' = e^{- t} + 0.07 - t e^{- t}$ , giving $A' = 10000 e^{h (t)} h' (t)$ .
Why: Both the time factor and the changing rate contribute through the product rule.
At t=3, h≈0.359361 and h′≈−0.029574, so $A' \approx - 423.62$ .
Why: Use the full initial amount 10000 dollars and retain units of dollars/year.
If r instead denotes the instantaneous percentage return, then $A' = r A$ is positive because r>0; that is a different growth model.
Why: A varying exponent coefficient is not interchangeable with an instantaneous growth rate.

Reviewed answer: Conditional on $A (t) = 10000 e^{t (e^{- t} + 0.07)}$ , $A' (3) \approx - 423.62$ dollars/year.

Takeaway: A rate description must specify the model before differentiation.

Final answer

Conditional on

A (t) = 10000 e^{t (e^{- t} + 0.07)}

A' (3) \approx - 423.62

dollars/year.

Return to problem 1 →

Problem 2

Original problem · clarified · added worked solution

You need $50 000 per year to live comfortably when you retire. Your bank offers a high interest savings account that gives you a compounding interest of 5% per year. How much money do you need to invest now if you want to live comfortably off the interest during your retirement 30 years from now ?

Clarification: Assume continuous compounding. The retirement fund is chosen so that interest at 5% supplies $50,000 per year without using the principal.

Authored worked explanation for the original source problem.

Source review note: The source value matches continuous compounding. Annual discrete compounding would instead give 1000000/(1.05)^30≈231377.45; the wording does not explicitly choose a convention.

To obtain 50000 dollars/year at a 5% continuous rate, the retirement balance must satisfy $0.05 B = 50000$ , so B=1000000 dollars.
Why: Divide desired yearly income by the specified rate.
If P is today's investment, $P e^{0.05 \cdot 30} = 1000000$ , so $P = 1000000 e^{- 1.5}$ .
Why: Discount the future balance with the same continuous model.

Reviewed answer: $1000000 e^{- 1.5} \approx 223130.16$ dollars under the source’s continuous model.

Takeaway: Use a consistent compounding convention throughout the calculation.

Final answer

1000000 e^{- 1.5} \approx 223130.16

dollars under the source’s continuous model.

Return to problem 2 →

Problem 3

Original problem · added worked solution

You want to maximize revenues for your company by changing the price. Your sales follow the function

S = 8 - 3 \ln P

, where S is the quantities sold (in 10 000 units) and P is the price. You want to keep the price of your product between $1 and $12.
a) What is the optimum price point?
b) How many units will you sell at this price?

Authored worked explanation for the original source problem.

Source review note: The source price 5.30 dollars is a coarse approximation; e^(5/3)≈5.29449 rounds to 5.29 dollars to the nearest cent. The source sales quantity is consistent.

Revenue in 10000-dollar units is $R (P) = P (8 - 3 \ln P)$ on [1,12].
Why: Revenue equals price times quantity.
$R' = 5 - 3 \ln P = 0$ gives $P = e^{5 / 3}$ ; $R'' = - 3 / P < 0$ .
Why: A negative second derivative makes this critical point a strict maximum.
R′ is positive below that price and negative above it, so it also beats both endpoints. At that price S=8−5=3, or 30000 units.
Why: A constrained absolute optimum needs its feasible interval checked.

Reviewed answer: $P = e^{5 / 3} \approx 5.2945$ dollars; sales 30000 units.

Takeaway: Optimize revenue, then convert the sales scale to actual units.

Final answer

P = e^{5 / 3} \approx 5.2945

dollars; sales 30000 units.

Return to problem 3 →

Problem 4

Original problem · added worked solution

You want to maximize profits by changing the quantity of laptops your company produces. Each laptop brings in $1000 of revenue. The total cost of producing laptops per year is represented by

C = \frac{Q^{3}}{10,000}

, where C is the cost and Q is the quantity of laptops produced.
a) What is the optimum number of laptops your company should make for profit maximization?
b) What is the annual profit at that quantity?

Authored worked explanation for the original source problem.

Source review note: The source lists only 1,200,000 dollars, which is a coarse profit approximation and omits the requested production quantity.

Profit is $Π (Q) = 1000 Q - Q^{3} / 10000$ for Q≥0.
Why: Subtract total cost from revenue before differentiating.
$Π' = 1000 - 3 Q^{2} / 10000 = 0$ gives the positive critical quantity shown; $Π'' = - 6 Q / 10000 < 0$ .
Why: The negative quantity root is outside the production domain.
At that Q, $Q^{3} / 10000 = (1000 / 3) Q$ , so $Π = (2000 / 3) Q \approx 1217161.24$ .
Why: The critical-point equation simplifies the profit calculation.
Compare integer quantities 1825 and 1826; 1826 has the slightly larger profit.
Why: Laptops are indivisible, so the continuous result must be checked against neighboring integers.

Reviewed answer: $Q = \sqrt{10000000 / 3} \approx 1825.74$ laptops in the continuous model; maximum profit ≈1217161.24 dollars; integer optimum 1826 laptops.

Takeaway: A continuous optimum is a guide to the discrete production choice.

Final answer

Q = \sqrt{10000000 / 3} \approx 1825.74

laptops in the continuous model; maximum profit ≈1217161.24 dollars; integer optimum 1826 laptops.

Return to problem 4 →

Problem 5

Original problem · added worked solution

As the size of your company increases, its percentage growth decreases. However, your company is adding more employees per year than ever before! Explain these concepts using variables, derivatives, and related rates.

Authored worked explanation for the original source problem.

$N' = 200 (t + 1)$ employees/year grows with t, while $N' / N = 2 / (t + 1)$ per year falls.
Why: Absolute growth counts new employees; relative growth divides that count by the current workforce.
The larger workforce can therefore add more people while each year's increase is a smaller percentage of the total.
Why: The two statements compare different quantities, so they do not conflict.

Reviewed answer: Example: $N (t) = 100 (t + 1)^{2}$ for t≥0 has increasing $N'$ but decreasing $N' / N$ .

Takeaway: A falling percentage rate can coexist with a rising absolute rate.

Final answer

Example:

N (t) = 100 (t + 1)^{2}

for t≥0 has increasing

N'

but decreasing

N' / N

Return to problem 5 →

Problem 6

Original problem · added worked solution

What is the absolute maximum value of the graph of

y = 10 x - x^{2}

Authored worked explanation for the original source problem.

$y' = 10 - 2 x = 0$ gives x=5 and $y'' = - 2 < 0$ .
Why: The parabola opens downward.
Alternatively $y = 25 - (x - 5)^{2} \leq 25$ for every real x.
Why: Completing the square proves the maximum is global.

Reviewed answer: $25$ at x=5.

Takeaway: A global inequality can strengthen a local derivative test.

Final answer

25

at x=5.

Return to problem 6 →

Problem 7

Original problem · added worked solution

For the equation

y = 3 \ln x - \sqrt{x}

,
a) At which point does its absolute maximum value occur?
b) What is the maximum value?

Authored worked explanation for the original source problem.

For x>0, $y' = 3 / x - 1 / (2 \sqrt{x}) = (6 - \sqrt{x}) / (2 x)$ .
Why: A common denominator exposes the derivative's sign.
The derivative is positive for x<36 and negative for x>36, so 36 gives the absolute maximum.
Why: The function rises before and falls after this unique critical point.
Substitute into the original function to obtain 3ln36−6.
Why: The maximum value is a function height, not a derivative value.

Reviewed answer: At $x = 36$ , maximum $3 \ln 36 - 6 \approx 4.75056$ .

Takeaway: A sign change across the whole domain proves an absolute maximum.

Final answer

x = 36

, maximum

3 \ln 36 - 6 \approx 4.75056

Return to problem 7 →

Problem 8

Original problem · added worked solution

As you pump air into a spherical balloon at a rate of 3cm³ per second, its radius increases to 2 cm. What is the rate of change of its surface area?

Authored worked explanation for the original source problem.

Source review note: The source gives 4/3 cm²/s. With the stated volume rate 3 cm³/s and radius 2 cm, the reviewed surface rate is 3 cm²/s.

$V = (4 / 3) π r^{3}$ gives $V' = 4 π r^{2} r'$ ; at r=2 and V′=3, $r' = 3 / (16 π)$ cm/s.
Why: Differentiate with respect to time before inserting the instant's radius.
$A = 4 π r^{2}$ gives $A' = 8 π r r' = 8 π \cdot 2 \cdot 3 / (16 π) = 3$ cm²/s.
Why: Surface rate and radius rate are related but have different units.

Reviewed answer: $3$ cm²/s.

Takeaway: Keep the geometric relation through both rate equations.

Final answer

3

cm²/s.

Return to problem 8 →

Problem 9

Original problem · added worked solution

You have an inflatable cube that constantly increases in size at 1cm per second. What is the rate of change in its volume when one of its sides is 4 cm?

Authored worked explanation for the original source problem.

If side length is s, $V = s^{3}$ , so $V' = 3 s^{2} s'$ .
Why: Volume changes with both current side length and its growth rate.
At s=4 and s′=1, the rate is 3·16·1=48 cm³/s.
Why: Insert the current side after differentiation.

Reviewed answer: $48$ cm³/s.

Takeaway: A linear side rate produces a size-dependent volume rate.

Final answer

48

cm³/s.

Return to problem 9 →

Problem 10

Original problem · clarified · added worked solution

Your friend gets distracted while holding up your ladder and you feel yourself starting to fall. Assuming gravity causes the vertical displacement from your original position to be defined as

x = - 3 t^{2}

, at what rate is the angle between the ladder and the ground changing after 0.5s?

Clarification: Treat the ladder as rigid, with its foot sliding on horizontal ground. Use the source diagram’s initial height 5 m and base distance 3 m; the top height is 5−3t² metres.

Authored worked explanation for the original source problem.

Source review note: The result uses the source diagram’s 5 m height and 3 m base with a rigid ladder and horizontal ground; the source rounds it to −0.75 rad/s.

The source diagram starts with vertical height 5 m and base distance 3 m, so the fixed ladder length is $L = \sqrt{34}$ m.
Why: Use the initial right triangle to obtain the rigid hypotenuse.
At t=0.5, height is $h = 5 - 3 t^{2} = 17 / 4$ m and $h' = - 6 t = - 3$ m/s.
Why: The printed displacement is measured from the initial height.
Since h=L sinθ, $h' = L \cos θ \cdot θ'$ and $L \cos θ = \sqrt{L^{2} - h^{2}} = \sqrt{255 / 16}$ .
Why: The horizontal leg is the coefficient converting vertical speed into angular speed.
Divide to obtain θ′≈−0.75147 rad/s, which rounds to −0.75.
Why: The negative sign means the angle is decreasing.

Reviewed answer: $- 3 / \sqrt{255 / 16} \approx - 0.75147$ rad/s, assuming the rigid ladder geometry in the source image.

Takeaway: The ladder length stays fixed while both legs change.

Final answer

- 3 / \sqrt{255 / 16} \approx - 0.75147

rad/s, assuming the rigid ladder geometry in the source image.

Return to problem 10 →

Problem 11

Added problem and worked solution

Find and classify the local extrema of $f (x) = x^{3} - 3 x$ .

Authored solution to an added practice problem.

$f' = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$ , so the critical inputs are ±1.
Why: Stationary points occur where the derivative vanishes.
$f'' = 6 x$ is negative at −1 and positive at 1; $f (- 1) = 2$ and $f (1) = - 2$ .
Why: The second derivative classifies the curvature at each stationary point.

Reviewed answer: Local maximum at (−1,2); local minimum at (1,−2).

Takeaway: State both the critical input and its height.

Final answer

Local maximum at (−1,2); local minimum at (1,−2).

Return to problem 11 →

Problem 12

Added problem and worked solution

Find the absolute extrema of $f (x) = x^{2} + 2 x - 1$ on $[- 2, 2]$ .

Authored solution to an added practice problem.

$f' = 2 x + 2 = 0$ gives x=−1 in [−2,2].
Why: Interior critical points are candidates along with endpoints.
$f (- 2) = - 1$ , $f (- 1) = - 2$ , and $f (2) = 7$ .
Why: Evaluate the original function at every candidate.
The smallest candidate value is −2 and the largest 7.
Why: Continuity on the closed interval guarantees the extrema are attained.

Reviewed answer: Absolute minimum −2 at x=−1; absolute maximum 7 at x=2.

Takeaway: Endpoints can win even when they are not stationary points.

Final answer

Absolute minimum −2 at x=−1; absolute maximum 7 at x=2.

Return to problem 12 →

Problem 13

Added problem and worked solution

A student says every point with f′=0 is a maximum or minimum. Use $f (x) = x^{3}$ to test the claim.

Authored solution to an added practice problem.

For $f = x^{3}$ , f′=3x²≥0 on both sides of zero.
Why: The derivative does not change from increasing to decreasing or vice versa.
The graph passes from negative heights to positive heights through zero.
Why: Values on either side lie on opposite sides of f(0), so it is neither a maximum nor a minimum.

Reviewed answer: No local maximum or minimum at zero.

Takeaway: A zero derivative is only a candidate for an extremum.

Final answer

No local maximum or minimum at zero.

Return to problem 13 →

Problem 14

Added problem and worked solution

A rectangular garden uses 40 m of fence on all four sides. Find the largest possible area and dimensions.

Authored solution to an added practice problem.

For sides x and y, $2 x + 2 y = 40$ gives y=20−x.
Why: The perimeter is the constraint.
$A = x (20 - x)$ has $A' = 20 - 2 x = 0$ at x=10, with $A'' = - 2$ .
Why: Area is the quantity being maximized.
Then y=10; endpoints of the feasible interval give zero area.
Why: The positive critical point beats degenerate endpoint rectangles.

Reviewed answer: A 10 m by 10 m square; maximum area 100 m².

Takeaway: Use the constraint to reduce to one variable.

Final answer

A 10 m by 10 m square; maximum area 100 m².

Return to problem 14 →

Problem 15

Added problem and worked solution

A river forms one side of a rectangular garden. With 40 m of fence for the other three sides, find the dimensions maximizing area.

Authored solution to an added practice problem.

If x is each side perpendicular to the river and y is the parallel side, $2 x + y = 40$ .
Why: Only three sides require fence.
$A = x (40 - 2 x)$ has derivative 40−4x, so x=10 and y=20.
Why: The negative second derivative proves the interior maximum.
The maximum area is 10·20=200 m².
Why: Use the actual dimensions to compute the objective.

Reviewed answer: Widths 10 m and length 20 m; area 200 m².

Takeaway: Draw the constraint before differentiating.

Final answer

Widths 10 m and length 20 m; area 200 m².

Return to problem 15 →

Problem 16

Added problem and worked solution

Cut equal squares from the corners of a 12 cm square sheet and fold an open box. Find the cut size maximizing volume.

Authored solution to an added practice problem.

$V (x) = x (12 - 2 x)^{2}$ for 0≤x≤6.
Why: Cutting removes twice x from each sheet dimension.
$V' = (12 - 2 x) (12 - 6 x)$ vanishes at x=2 and the degenerate endpoint 6.
Why: Use the product and chain rules.
V(0)=V(6)=0 and V(2)=128, so the useful maximum occurs at 2.
Why: The closed feasible interval includes the endpoint comparisons.

Reviewed answer: Cut squares of side 2 cm; box 8×8×2 cm; volume 128 cm³.

Takeaway: An endpoint critical value can describe a collapsed box.

Final answer

Cut squares of side 2 cm; box 8×8×2 cm; volume 128 cm³.

Return to problem 16 →

Problem 17

Added problem and worked solution

A closed cylindrical can holds $16 π$ cm³. Find the radius and height minimizing its surface area.

Authored solution to an added practice problem.

With $V = 16 π$ , the constraint gives $h = 16 / r^{2}$ .
Why: Volume remains fixed while dimensions vary.
Surface area $A = 2 π r^{2} + 2 π r h = 2 π r^{2} + 32 π / r$ has $A' = 4 π r - 32 π / r^{2}$ .
Why: Include both circular ends for a closed cylinder.
$A' = 0$ gives r³=8, so r=2 and h=4. $A'' = 4 π + 64 π / r^{3} > 0$ .
Why: Strict positive curvature gives the unique global minimum on r>0.

Reviewed answer: Radius 2 cm and height 4 cm.

Takeaway: Closed and open cylinders have different area objectives.

Final answer

Radius 2 cm and height 4 cm.

Return to problem 17 →

Problem 18

Added problem and worked solution

Find the point on $y = x$ closest to $(2, 0)$ .

Authored solution to an added practice problem.

A point on y=x is (x,x). Squared distance to (2,0) is $D (x) = (x - 2)^{2} + x^{2}$ .
Why: Minimizing distance and its nonnegative square gives the same point.
$D' = 4 x - 4 = 0$ gives x=1 and $D'' = 4 > 0$ .
Why: The squared distance is a convex quadratic.
The point is (1,1), at distance √2.
Why: Take the square root only after locating the optimum.

Reviewed answer: The closest point is (1,1); distance $\sqrt{2}$ .

Takeaway: A squared-distance objective avoids a needless radical derivative.

Final answer

The closest point is (1,1); distance

\sqrt{2}

Return to problem 18 →

Problem 19

Added problem and worked solution

A fictional shop sells $q = 200 - 10 p$ units at price p dollars. On $0 \leq p \leq 20$ , find the price maximizing revenue.

Authored solution to an added practice problem.

$R (p) = p (200 - 10 p)$ for 0≤p≤20.
Why: The upper price keeps the modeled sales nonnegative.
$R' = 200 - 20 p = 0$ at p=10; $R'' = - 20 < 0$ .
Why: The quadratic has a unique maximum.
At p=10, q=100, R=1000; both endpoints give zero.
Why: Verify the constraint and evaluate actual units.

Reviewed answer: Price $p = 10$ dollars; sales 100 units; maximum revenue $1000$ .

Takeaway: A sales model needs a feasible price interval.

Final answer

Price

p = 10

dollars; sales 100 units; maximum revenue

1000

Return to problem 19 →

Problem 20

Added problem and worked solution

A circular ripple has radius 3 cm growing at 2 cm/s. How fast is its area changing?

Authored solution to an added practice problem.

$A = π r^{2}$ gives $A' = 2 π r r'$ .
Why: Area depends on the changing radius.
At r=3 cm and r′=2 cm/s, A′=12π cm²/s.
Why: Keep the length units in both factors.

Reviewed answer: $12 π$ cm²/s.

Takeaway: Differentiate before substituting the instant's radius.

Final answer

12 π

cm²/s.

Return to problem 20 →

Problem 21

Added problem and worked solution

A sphere gains volume at $8 π$ cm³/s when its radius is 2 cm. Find its radius rate.

Authored solution to an added practice problem.

$V' = 4 π r^{2} r'$ .
Why: Differentiate the sphere volume with respect to time.
At V′=8π and r=2, $r' = 8 π / (16 π) = 1 / 2$ cm/s.
Why: Solve for the requested linear rate rather than the volume rate.

Reviewed answer: $1 / 2$ cm/s.

Takeaway: The current surface area converts volume rate into radius rate.

Final answer

1 / 2

cm/s.

Return to problem 21 →

Problem 22

Added problem and worked solution

A 5 m ladder rests on level ground against a wall. Its base moves away at 2 m/s. How fast does the top fall when the base is 3 m from the wall?

Authored solution to an added practice problem.

A 5 m ladder satisfies $x^{2} + y^{2} = 25$ , so $2 x x' + 2 y y' = 0$ .
Why: The fixed length provides the relation between the two legs.
At x=3, y=4 and x′=2, so $y' = - x x' / y = - 6 / 4$ .
Why: The height falls while the base moves outward.

Reviewed answer: $y' = - 3 / 2$ m/s.

Takeaway: The sign carries the physical direction.

Final answer

y' = - 3 / 2

m/s.

Return to problem 22 →

Problem 23

Added problem and worked solution

A cone-shaped tank has height 6 m and top radius 3 m, point downward. Water enters at 1 m³/min. Find the height rate when the water is 2 m deep.

Authored solution to an added practice problem.

Similarity gives r/h=3/6=1/2, so $V = (1 / 3) π (h / 2)^{2} h = π h^{3} / 12$ .
Why: The water surface radius changes with height; it is not the full tank radius.
$V' = π h^{2} h' / 4$ . At h=2 and V′=1, h′=1/π.
Why: Insert the current water depth after differentiation.

Reviewed answer: Water height rises at $1 / π$ m/min.

Takeaway: Use similar triangles to eliminate the changing radius.

Final answer

Water height rises at

1 / π

m/min.

Return to problem 23 →

Problem 24

Added problem and worked solution

Two walkers leave the same point at the same time. One goes east at 3 m/s and one north at 4 m/s. How fast does their separation grow?

Authored solution to an added practice problem.

If one walker is x meters east and the other y meters north of their common start, $d^{2} = x^{2} + y^{2}$ .
Why: The separation is a hypotenuse.
At the instant x=3t, y=4t, so d=5t and d′=5 m/s for t>0.
Why: The constant speeds form a 3–4–5 right triangle at every time.

Reviewed answer: $5$ m/s.

Takeaway: Two perpendicular speeds combine through geometry, not simple addition.

Final answer

5

m/s.

Return to problem 24 →

Problem 25

Added problem and worked solution

A rectangle is 5 cm long and 4 cm wide. Its length shrinks at 1 cm/s while its width grows at 2 cm/s. Find its area rate.

Authored solution to an added practice problem.

For a rectangle, $A' = l w' + w l'$ .
Why: Both dimensions change, so both product contributions matter.
At l=5, w=4, l′=−1, w′=2, the area rate is 5·2+4(−1)=6 cm²/s.
Why: A shrinking length gives a negative contribution but the width's increase is larger.

Reviewed answer: $6$ cm²/s.

Takeaway: Competing changes need their signed contributions added.

Final answer

6

cm²/s.

Return to problem 25 →

Problem 26

Added problem and worked solution

A sign chart gives f′>0 on $(- \infty, - 2)$ , f′<0 on $(- 2, 1)$ , and f′>0 on $(1, \infty)$ . Assume f is continuous and stationary at −2 and 1. Classify the two points.

Authored solution to an added practice problem.

The derivative changes from positive to negative at −2.
Why: The function rises then falls, so its nearby values are lower than the height at −2.
At 1 it changes from negative to positive.
Why: The function falls then rises, creating a local minimum.

Reviewed answer: Local maximum at −2; local minimum at 1.

Takeaway: A first-derivative sign change classifies an extremum.

Final answer

Local maximum at −2; local minimum at 1.

Return to problem 26 →

Problem 27

Added problem and worked solution

Does $f (x) = x$ have absolute extrema on $(0, 1)$ ? Explain the role of the endpoints.

Authored solution to an added practice problem.

f(x)=x is increasing, and all its values lie strictly between 0 and 1.
Why: The interval omits both endpoints.
Any proposed smallest positive value can be halved to obtain a smaller allowed value; similarly any input below 1 has a larger allowed input.
Why: The bounding heights are approached but never attained.

Reviewed answer: No absolute minimum on (0,1); infimum 0. No absolute maximum either; supremum 1.

Takeaway: Bounds are not extrema unless the function actually reaches them.

Final answer

No absolute minimum on (0,1); infimum 0. No absolute maximum either; supremum 1.

Return to problem 27 →

Problem 28

Added problem and worked solution

Maximize $f (x) = 20 - (x - 5)^{2}$ on $[0, 3]$ .

Authored solution to an added practice problem.

$f' = - 2 (x - 5) > 0$ throughout [0,3].
Why: The unrestricted peak at x=5 is outside the allowed interval.
The function increases to the right endpoint, where $f (3) = 20 - 4 = 16$ .
Why: A constraint can make an endpoint the optimum.

Reviewed answer: Maximum at x=3 with value 16.

Takeaway: Only feasible points are candidates.

Final answer

Maximum at x=3 with value 16.

Return to problem 28 →

Problem 29

Added problem and worked solution

A fictional team has 400 employees and gains 12 employees/year at the current instant. State its absolute and percentage growth rates.

Authored solution to an added practice problem.

Divide the headcount rate 12 by the current headcount 400 to get 0.03 per year.
Why: A percentage rate is normalized by the current amount.
Multiply by 100 to express 0.03 as 3%.
Why: The absolute count and relative percentage use different units.

Reviewed answer: Absolute growth 12 employees/year; relative growth 3%/year.

Takeaway: Name the denominator when reporting a percentage rate.

Final answer

Absolute growth 12 employees/year; relative growth 3%/year.

Return to problem 29 →

Problem 30

Added problem and worked solution

Two positive numbers have product 36. Find the pair with the smallest sum.

Authored solution to an added practice problem.

For x>0 and xy=36, write $S = x + 36 / x$ .
Why: The product constraint eliminates y.
$S' = 1 - 36 / x^{2} = 0$ gives x=6; $S'' = 72 / x^{3} > 0$ .
Why: The sum is strictly convex on its positive domain.
Then y=6 and S=12; the sum grows without bound as x approaches 0 or infinity.
Why: This is the global minimum, not just a local one.

Reviewed answer: The two positive numbers are both 6; minimum sum 12.

Takeaway: Constraints and endpoint behavior turn a local test into a global conclusion.

Final answer

The two positive numbers are both 6; minimum sum 12.

Return to problem 30 →

Chapter answer key and worked steps

Answers and worked steps for the original problems, including verified corrections where needed.

Checked answer

Conditional on $A (t) = 10000 e^{t (e^{- t} + 0.07)}$ , $A' (3) \approx - 423.62$ dollars/year.
Problem 1 →
Checked answer

$1000000 e^{- 1.5} \approx 223130.16$ dollars under the source’s continuous model.
Problem 2 →
Checked answer

$P = e^{5 / 3} \approx 5.2945$ dollars; sales 30000 units.
Problem 3 →
Checked answer

$Q = \sqrt{10000000 / 3} \approx 1825.74$ laptops in the continuous model; maximum profit ≈1217161.24 dollars; integer optimum 1826 laptops.
Problem 4 →
Original short answer

Answers will vary
Problem 5 →
Original short answer

$25$
Problem 6 →
Original short answer

a. $36$
b. $4.75$
Problem 7 →
Corrected answer

$3$ cm²/s.
Problem 8 →
Original short answer

$\frac{d V}{d t} = 48 \frac{{cm}^{3}}{s}$
Problem 9 →
Checked answer

$- 3 / \sqrt{255 / 16} \approx - 0.75147$ rad/s, assuming the rigid ladder geometry in the source image.
Problem 10 →

Read the chapter answer key in the book →

Source worked examples

These are the book’s example problems and worked steps, with the original explanations.

Example Problem 6.1.1

Maximinima →

Find the absolute maximinima values and where they occur of the function $f (x) = x^{3} - 6 x^{2} + 9 x - 2$ on the closed interval $[0,4]$ .

Our first step is to find the points where the derivative equals zero.

$\begin{matrix} f^{'} (x) = 3 x^{2} - 12 x + 9 \\ = 3 (x 2 - 4 x + 3) \\ = 3 (x - 3) (x - 1) \\ x = 3, x = 1 \end{matrix}$

Our two interior critical points, x = 1 and x = 3, lie in the interval. We also test the endpoints x = 0 and x = 4. These four candidates—not four interior critical points—are the values to compare for absolute extrema.

$f (0) = 0 - 0 - 0 - 2 = - 2$

$f (1) = 1 - 6 + 9 - 2 = 2$

$f (3) = 27 - 54 + 27 - 2 = - 2$

$f (4) = 64 - 96 + 36 - 2 = 2$

We can see from our calculations that $f (0)$ and $f (3)$ are both absolute minimums since they share the same value. Similarly, $f (1)$ and $f (4)$ are both absolute maximums. This is why it is important to take boundary conditions into account!

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Example Problem 6.1.2

Maximinima →

Find and label all the critical x-values with either “local maxima” or “local minima” of the function $f (x) = \sin (x^{2})$ on the closed interval $[0, π]$ .

As always, we find the derivative first

$f^{'} (x) = 2 x \cdot \cos (x^{2})$

The derivative equals zero when $2 x = 0$ or $\cos (x^{2}) = 0$ . We know from our unit circle that $\cos (x) = 0$ when $x = (2 n + 1) \cdot π / 2$ , with n being whole numbers $\geq 0$ . Therefore, $\cos (x^{2}) = 0$ on the square root of these intervals, but how many n’s do we count?

Well, we know that the upper limit of our interval $[0, π]$ is $x = π$ , so we stop before we hit $\cos (π^{2})$ . Let’s do some quick mental math to see how many critical points we are working with. Notice that after x = 0, the 2x term in the derivative has no effect on the critical points.

$\begin{matrix} 0 \leq \sqrt{\frac{π}{2}} \leq \sqrt{π^{2}} \\ 0 \leq \sqrt{\frac{3 π}{2}} \leq \sqrt{π^{2}} \\ 0 \leq \sqrt{\frac{5 π}{2}} \leq \sqrt{π^{2}} \\ 0 \leq \sqrt{\frac{7 π}{2}} \leq \sqrt{π^{2}} \end{matrix}$

So we have about 3 critical points, plus 0, which makes 4. $π$ does not count because $\cos (π^{2}) \neq 0.$

So which ones are local maximum and which ones are local minimum? Let’s take the second derivative and find out.

$f^{''} (x) = 2 \cos (x^{2}) - 4 x^{2} \sin (x^{2})$

This may look daunting, and you might be thinking, “let’s just plug the critical points into the original function” but remember, we only need to know whether the second derivative is positive or negative. We do not need to solve for its value, so we can estimate. Let’s start with x = 0, the easiest.

$f^{''} (0) = 2 \cdot 1 - 0 =$ Positive (Concave up. We are at a minimum)

$f^{''} (\sqrt{\frac{π}{2}}) = 0 -$ Big Number

(We get a large negative number. Concave down. We are at a maximum)

Notice that $\cos (x^{2})$ will always be 0 at the critical points because that is the derivative, so for the second derivative we only need to worry about the sine term, whether it is positive or negative.

$f^{''} (\sqrt{\frac{3 π}{2}}) = 0 -$ Big Negative Number

$(\frac{3 π}{2}$ is between quadrants III and IV, where the sine function is negative. Two negatives make a positive! This means concave up. We are at a minimum).

$f^{''} (\sqrt{\frac{5 π}{2}}) = 0 -$ Big Positive Number

(Subtracting a big positive number from 0 gives a big negative number. Concave down. We are at a maximum).

So in summary:

$f (0) =$ local min

$f (\sqrt{\frac{π}{2}}) =$ local max

$f (\sqrt{\frac{3 π}{2}}) =$ local min

$f (\sqrt{\frac{5 π}{2}}) =$ local max

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Example Problem 6.2.1

Optimization →

You really want to win that contest! So you decide to start working on the highly efficient can that holds 250 ${cm}^{3}$ of soda using the least amount of aluminum.

a) Find the dimensions of the cylindrical can.
b) What is the cost to manufacture one can, assuming a sheet of aluminum costs $0.001 per cm²?
c) Is this design realistic? If not, what should the dimensions be, given that the radius can’t be wider than 10 cm and the height no longer than 20 cm?

a) You might be thinking that we are not given enough information. Let’s write down the two equations we know related to cylinders.

$A = 2 π r h + 2 π r^{2}$

$V = π r^{2} h$

We don’t know the area, radius, or height, but we do know the volume, which is 250 ${cm}^{3}$ . In all optimization problems, there are always at least two of the following variables:

The optimizing variable - This is the variable to be maximized or minimized. It could be profits, volume, area, perimeter, anything.

2. The constraint - This is our limiting variable. This is something that is fixed that will not change in the problem. Usually we are given the value of constraints. What is our constraint in this problem? Volume? Area? Radius? Height?

If you think Area is the constraint... *buzz* so sorry, but no. Area is the optimizing variable, because we want to use as little aluminum as possible. If you picked volume as the constraint, you would be correct, because we are given how many milliliters of soda each can should hold. This is not going to change. Remember, we want to minimize area, and keep the volume constant. What do you think the derivative of the area equation is going to be?

$250 = π r^{2} h$

We do not want to have to worry about two variables in our area equation, so let’s find h in terms of r by dividing both sides by $π r^{2}$ .

$\frac{250}{π r^{2}} = h$

Now let’s plug this into the Area equation, then take the derivative.

$A = 2 π r \frac{250}{π r^{2}} + 2 π r^{2}$

$A = \frac{500}{r} + 2 π r^{2}$
Now we can find the derivative of Area and set it equal to zero because this is the quantity we want to minimize. Notice that by setting h in terms of r, we are only differentiating using 1 variable, and do not have to worry about implicit differentiation.

$A^{'} = - \frac{500}{r^{2}} + 4 π r$

$0 = - \frac{500}{r^{2}} + 4 π r$
$\frac{500}{r^{2}} = 4 π r$
$r^{3} = \frac{500}{4 π}$
$r = \sqrt[3]{39.8 {cm}^{3}} = 3.4 cm$

The question asks us to find the dimensions of the can, so let’s knock out height and Area.

$\frac{250}{π (3.4)^{2}} = h$

$h = 6.9 cm$
$A = 2 π (3.4) (6.9) + 2 π (3.4)^{2}$

$A = 220 {cm}^{2}$

The cost of aluminum is given as a tenth of a penny per ${cm}^{2}$ . Our can has an area of 220 ${cm}^{2}$ . This is simple math.

$cost per can = 220 {cm}^{2} \cdot \frac{$ 0.001}{{cm}^{2}} = $ 0.22$

Reality check: See the following image of the proposed can (interactive 6.2.1).

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Example Problem 6.3.1

Related Rates →

A droplet forming inside of a cloud has a diameter of 0.1 mm. The high humidity inside the cloud causes the rate of condensation to be 1 mm³/s.
a) At what rate does the diameter of the droplet increase after 3 seconds?

b) What was the original rate of increase of the droplet size when its diameter was still 0.1 mm?

Here we are going to assume that the initial volume is negligible and the only thing that matters is the volume 3 seconds after condensation starts to occur. Why? because: $V = \frac{4}{3} (0.1)^{3} = 0.0042 {mm}^{3} ≪ 3 {mm}^{3}$
In other words, the difference in volume is less than 0.2%. So let’s assume our droplet’s current volume is 3 ${mm}^{3}$ . You are going to love what happens next! First, here is what we know:

$V = \frac{4}{3} π r^{3}$

$\frac{d V}{d t} = \frac{1 {mm}^{3}}{s}$

$\frac{d r}{d t} = unknown$

Can you guess what we have to do? Starts with an I. Rhymes with differentiation.

That’s right! Implicit differentiation! How? Let’s take the general derivative with respect to time of the volume function.

$\frac{d}{d t} V = \frac{d}{d t} (\frac{4}{3} π r^{3})$
$\frac{d V}{d t} = \frac{4}{3} π \cdot (\frac{d}{d r} r^{3}) \cdot \frac{d r}{d t}$

Now we can take the derivative with respect to r of the term inside the parentheses.

$\frac{d V}{d t} = \frac{4}{3} π \cdot (3 r^{2}) \cdot \frac{d r}{d t}$

$\frac{d V}{d t} = \frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t}$

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

We know two of these unknown variables. We can solve for r after 3 seconds, and find $\frac{d r}{d t}$ . First let’s solve for r for the new volume after 3 seconds.

$3 {mm}^{3} = \frac{4}{3} π r^{3}$

$r = \sqrt[3]{0} .716 = 0.895 mm$

What does this tell us? We can see that in just 3 seconds, the size of the droplet increased 9 times!

However, we know the rate slows down the larger the droplet gets. How much slower? We are about to find out (bolded). Let’s plug in r into our related rates equation.

$\frac{1 {mm}^{3}}{s} = 4 π (0.895 mm)^{2} \cdot \frac{d r}{d t}$

$\frac{1 {mm}^{3}}{4 π \cdot 0.801} = \frac{d r}{d t}$

$\frac{d r}{d t} = 0.1 \frac{mm}{s}$

Are we done? No! The question asked us for the rate of diameter increase. Common sense tells us this is just twice the rate of radial increase, but we can prove it mathematically too, using implicit differentiation!

$2 r = D$

$\frac{d}{d t} 2 r = \frac{d}{d t} D$

$2 \frac{d r}{d t} = \frac{d D}{d t}$

$\frac{d D}{d t} = 0.2 \frac{m m}{s}$

b) This part asks us to find what the rate of increase in diameter originally was. We have done most of the work already, so let’s just go back to our related rates equation and plug in a new value for the radius, 0.05 mm (this is half the original diameter given to us in the question).

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

$1 \frac{{mm}^{3}}{s} = 4 π (0.05 {mm}^{2}) \frac{d r}{d t}$

$\frac{1 \frac{mm}{s}}{4 π \cdot 0.0025} = \frac{d r}{d t}$

$\frac{d r}{d t} = 31.83 \frac{mm}{s}$

Does this number seem high...? Nope! The droplet really does increase at a size of 31.83 millimeters per second! Why? Because when the droplet is really small, any moderate increase in volume will cause its size to become 3x or even 4x larger! Once the droplet becomes bigger, the same amount of volume ends up having a smaller effect.

For example, if you were living on $1 a day, and someone gave you one dollar, you would be relatively much richer than if you were making $100 per day and someone gave you that same dollar. The same concept applies here. Do we have our answer yet? Nope! Because we are asked for the change in diameter.

$\frac{d D}{d t} = 2 \frac{d r}{d t} = 63.7 \frac{m m}{s}$

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Example Problem 6.3.2

Related Rates →

Your friend wants you to time him to see how fast he can run. He starts out 10 m east of you, and after you hit the start button on the timer, he runs at a speed of 5 m/s north. How fast is the angle between you changing relative to the original starting point after 3 seconds? (See Interactive 6.3.1 at left).

We know that after three seconds, your friend has gone north 15 m based on his constant velocity of 5 m/s. Now we have a right triangle with the adjacent side equal to 10, and the opposite side equal to 15. We can find θ.

$θ = \tan^{- 1} \frac{15}{10} = 56.3 º$

Now how do we find the related rates equation? Let’s look at which two variables are changing. θ is changing and the vertical distance (or opposite side) is changing. Do we know an equation that relates these two? Yes we do!

$\tan θ =$ $\frac{opposite}{adjacent} = \frac{N}{E}$

Rather than having to say opposite and adjacent all the time, let’s just call these north and east, because that is what they are. Since East doesn’t change, we can treat that as a coefficient to our changing variable, North, and take the implicit derivative with respect to time for both sides of the equation. (sizes below)

$\frac{d}{d t} \tan θ = \frac{1}{E} \cdot \frac{d}{d t} (N)$

$\sec^{2} (θ) \cdot \frac{d θ}{d t} = \frac{1}{E} \frac{d N}{d t}$

We know all of these values except for $\frac{d θ}{d t}$ , which we are trying to solve for. Notice that $\frac{d N}{d t}$ is just the velocity of your friend running north.

Plugging in these values gives us

$3.25 (\frac{d θ}{d t}) = \frac{1}{10 m} \cdot 5 m$

$\frac{d θ}{d t} = \frac{0.154}{s}$

Our answer is not in degrees, however! It is in radians. Degrees is a unit. Radians are unit-less. We can convert this to degrees by multiplying our answer by $\frac{180^{º}}{π}$ , which gives

$\frac{d θ}{d t} = \frac{8.82 º}{s}$

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Example Problem 6.4.1

FFR - Estimating Areas →

Using the trapezoids method, estimate the area under the curve of $y = \sin (π x)$ from x = 0 to x = 2, with n, the number of trapezoids, equal to 4.

Remember the area of a trapezoid is $h \cdot \frac{b_{1} + b_{2}}{2}$ . Since I am so generous, and this is the first problem, I will provide a picture as a reference. Think of this as training wheels. The next chapter gets real.

Take a few moments to understand the concepts highlighted in Interactive Image 6.4.2. We can then see that the areas are symmetric. Not only that, they cancel each other out (the blue cancels out the red)! So our total area is zero! You might ask, “how”?

Think about it this way. You start out at the entrance of your bedroom, do some vacuuming, rearrange your desk, and come back to the entrance. What is your net displacement? Zero! You ended up in the same place you started. However, what is your total distance traveled? Ah, now that would be a different answer.

Here, we are calculating net area, which is zero. What is the net area from x = 0 to x = 1? We can see that it is the sum of the area of the blue triangles, which is 0.5. What about the net area from x = 1 to x = 2? It is the sum of the red triangles, but they are under the x-axis, so the net area is -0.5! Adding these two up gives us a total area of 0.

Gallery 6.4.1 Estimation accuracy using different methods of Approximation

Read all original image captions

Image 1

Left-endpoint sums underestimate the definite integral of an increasing function and overestimate it for a decreasing function. The rectangles in these sketches show exactly that direction. Concavity does not change this monotonicity comparison.

Image 2

Increasing graphs have their area overestimated using the right corner point approximation method. Decreasing graphs have their area underestimated.

Image 3

For a function that is concave up throughout the interval, midpoint sums underestimate the definite integral. Concave down throughout gives an overestimate. The sketches show the direction of the error; the word “slight” in an image is not a general error bound. The size depends on the function and partition.

Image 4

Similar to the midpoint method, the trapezoidal method overestimates or underestimates based on concavity, not whether a graph is increasing or decreasing. Concave up graphs have their areas overestimated. Concave down graphs are underestimated.

Interactive 6.4.1 The best way to estimate

Magnification 1×

Read all original figure notes

Rectangle width Δx

Use n equal-width rectangles on the fixed interval [a,b]. Each width is Δx = (b − a)/n, and right endpoints give the sum Δx[f(a + Δx) + ⋯ + f(b)]. As n increases without bound, Δx approaches 0. For a continuous function, left, right, and midpoint sums converge to the same definite integral. An undefined endpoint does not by itself make an improper integral safe: that case needs its own limit.

As Δx approaches 0

Make the rectangles narrower while keeping [a,b] fixed. The number n tends to infinity, and their width Δx tends to 0. For a continuous function, the limiting signed sum is the definite integral. At a finite n, more rectangles can improve an estimate, but no universal comparison makes every individual estimate closer.

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Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 5

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Problem 6

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Problem 8

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Chapter answer key and worked steps

Source worked examples

Example Problem 6.1.1

Example Problem 6.1.2

Example Problem 6.2.1

Example Problem 6.3.1

Interactive 6.3.1 Example Problem 6.3.2

Finding θ

t = 3 s

Example Problem 6.3.2

Interactive 6.3.1 Example Problem 6.3.2

Finding θ

t = 3 s

Example Problem 6.4.1

Gallery 6.4.1 Estimation accuracy using different methods of Approximation

Image 1

Image 2

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Image 4

Interactive 6.4.1 The best way to estimate