Best Price

CED topics

• 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
• 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
• 5.7 Using the Second Derivative Test to Determine Extrema
• 5.10 Introduction to Optimization Problems
• 5.11 Solving Optimization Problems

Big idea and practices

Big idea: the derivative finds the best. A quantity that rises, then falls peaks at a critical point or at the edge of what the story allows. CED FUN-4.A (justify from derivatives) and FUN-4.B (applied extrema); FUN-1's Extreme Value Theorem backs the candidates test.

Students leave able to (practices in parentheses):

• find and classify critical points by the first or second derivative test (1.E, 3.D)
• find absolute extrema on a closed interval, endpoints included (3.D)
• set up optimization from a story: variable, function, domain; five stories, one downward parabola (2.A)
• explain why the seat cap moves the answer to an endpoint (3.E)
• answer in context with units, the slope's included; tell revenue from profit (3.F, 4.B)

45-minute plan

1. 0–5 minRead both hooks; pairs read "Raise or lower?" together and pick in it (or circle on the printout). Vote: raise or lower?
2. 5–12 minChallenges 1 and 2 in pairs, any method: table, vertex, or derivative. Board the whole-dollar table.
3. 12–18 minRevenue explorable (projected or on devices) and its four “Try these.”
4. 18–26 minRules, whole class: critical points, the three tests, the six steps. Label ${R'(p)}$ in ${\tfrac{\text{dollars of revenue}}{\text{dollar of price}}}$.
5. 26–30 minWorked example together. Challenge 3 (car wash) is homework.
6. 30–37 minChallenge 4 (seat cap). Pairs read the seat-cap thread together, both answering each question before tapping Next; then discuss.
7. 37–41 minChallenge 5 in pairs, then the “Revenue isn’t profit” callout.
8. 41–45 minExit ticket, then the reflection.

Homework: Challenge 3, Challenge 6 with the profit curve, fluency, mixed review.

Launch questions

• A ticket price goes up. How does that bring in more money? Less?
• Revenue at $0? At $1,000? And between?
• After predictions: is the most-revenue price the most-profit price? What would make them differ?
• The rink sells out every night. Raise the price?

Misconceptions and how to respond

• Whole dollars only. A whole-dollar table ties $11 and $12 at $7,920; on Challenge 1 the common misses are 11 and 12. Prices go to the cent.
• Answering a different question. Watch for 7935 on Challenge 1 and 11.5 on Challenge 2: revenue for price, and the reverse. Underline what's asked.
• The formula's critical point is always the answer. On Challenge 4, $10 zeros the derivative but draws 780 fans to 750 seats. Below $10.50 the rink sells out, so a raise only adds money; at $10.50 and up everyone fits and revenue falls. Best: $10.50, a sellout, for $10,125, at a corner of the revenue graph, where the derivative doesn't exist.
• Revenue is profit. Watch for 17.5 on Challenge 5, 900 on Challenge 6, and 35 on the exit ticket. Profit subtracts each unit's cost.
• Relative means absolute. On Fluency 6, 4 is the relative minimum value; the endpoint gives 0. Compare every candidate.
• A zero derivative means an extremum. On Fluency 2, ${x = -1}$ is critical, but ${f'}$ doesn't change sign. Sketch ${x^3}$ at 0.

Support and extension

Support

• Whole-dollar table, $8 to $13: $7,200, $7,560, $7,800, $7,920, $7,920, and $7,800. Where next, and why?
• Six steps on the board; students check each off on every problem.
• Functions in words first: revenue = price × fans, profit = (price − cost) × sales.
• Pairs: one runs the explorable, the other predicts the slope's sign before each move.

Extension

• With linear demand, the best profit price is the best revenue price plus half the unit cost. Phone cases: $17.50 plus half of $5 is $20. At a $7 cost? ($21.)
• Concessions, no seat cap: best price $10. How many seats make $10 the answer? (780 seats.)
• Hoodies at $14 to make instead of $12: best price? ($42.)
• Show that for ${R(p) = p(a - bp)}$, the best price draws half the fans a free ticket would.

Without devices

• Print the lesson as the handout: no header, buttons, or hints; room to write. Students circle their predictions.
• Explorable: plot the support table and sketch the parabola (by symmetry, $15 brings in the same $7,200 as $8); lay rulers on it at $8, $11.50, and $15. The slopes are 420, 0, and −420 dollars of revenue per dollar of price.
• Answer key below: the write-ups students unlock.
• Exit ticket on paper.

Answer key

Every problem in the order the lesson gives them, with the answer the page accepts and my full write-up. Answer boxes check only the final number, so ask students for their work.

1. Challenge 1

   No calculator · CED 5.10, 5.11

   At $8 a ticket, 900 fans want tickets. Every 10¢ increase loses 6 fans. The number of fans is a linear function of the price, and the price can be any amount to the cent. What price maximizes ticket revenue?

   Answer: $11.50

   1. Let ${p}$ be the price in dollars. $1 more is ten 10¢ steps: ${10 \times 6 = 60}$ fewer fans. Fans who want tickets: $\displaystyle n(p) = 900 - 60(p - 8)$$n(p)$$=900-60(p-8)$$\displaystyle = 1{,}380 - 60p$$=1,380-60p$multiply out
   2. Revenue in dollars is price times fans: $\displaystyle R(p) = p(1{,}380 - 60p)$$R(p)$$=p(1,380-60p)$$\displaystyle = 1{,}380p - 60p^2$$=1,380p-60p^2$multiply out
   3. ${R'(p) = 1{,}380 - 120p}$ dollars of revenue per dollar of price; it is 0 at ${p = 11.5}$.
   4. ${R'(p) > 0}$ for ${p < 11.5}$ and ${R'(p) < 0}$ for ${p > 11.5}$, so revenue rises on ${[0, 11.5]}$ and falls on ${[11.5, 23]}$ (no one buys at $23). ${R(11.5)}$ is the absolute maximum, at the parabola's vertex ${p = -\frac{1{,}380}{2(-60)}}$. The best ticket price is $11.50.

   Common wrong answers

   • $11.00: $11 and $12 tie at $7,920. Prices go to the cent: look between them.
   • $12.00: $11 and $12 tie at $7,920. Prices go to the cent: look between them.
   • $8.00: That's today's price. A raise loses fans but earns more per ticket.
   • $7,935.00: That's the most revenue. The question asks for the price.

2. Challenge 2

   No calculator · CED 5.11

   Same team. What is the most ticket revenue per game?

   Answer: $7,935.00

   1. At the best price, ${p = 11.5}$, ${1{,}380 - 60(11.5) = 690}$ fans buy tickets.
   2. Revenue in dollars: $\displaystyle R(11.5) = 11.5 \times 690$$R(11.5)$$=11.5\times 690$$\displaystyle = \boxed{7{,}935}$$=\boxed{7,935}$ The most is $7,935 a game.

   Common wrong answers

   • 690: That's the fans at the best price. Revenue is price times fans.
   • $11.50: That's the best price. The question asks for the revenue.
   • $7,200.00: That's revenue at today's $8. The best price brings in more.

3. Worked example

   No calculator · CED 5.5, 5.7, 5.10, 5.11

   A band sells 50 T-shirts a night at $20 each, and each $1 increase loses 2 sales. Sales are a linear function of the price, and the price can be any amount to the cent. What price maximizes T-shirt revenue?

   Worked on the page: $22.50

   1. Variable and units: ${p}$ is the price in dollars. Sales drop 2 per dollar above $20, so shirts sold are $\displaystyle q(p) = 50 - 2(p - 20)$$q(p)$$=50-2(p-20)$$\displaystyle = 90 - 2p$$=90-2p$multiply out
   2. Function: revenue in dollars is price times shirts sold: $\displaystyle R(p) = \underbrace{p}_{\text{price}} \cdot \underbrace{(90 - 2p)}_{\text{shirts sold}}$$R(p)$$=\underset{\text{price}}{\underbrace{p}}\cdot \underset{\text{shirts sold}}{\underbrace{(90-2p)}}$$\displaystyle = 90p - 2p^2$$=90p-2p^2$multiply out
   3. Domain: price and sales can't be negative, so ${0 \le p \le 45}$.
   4. Critical points: the tangent's slope ${\teal{R'(p)}}$ is 0 where the tangent is flat: $\displaystyle \teal{R'(p)} = 90 - 4p$$R^{\prime }(p)$$=90-4p$differentiate$\displaystyle 90 - 4p = 0$$90-4p$$=0$set the slope to 0$\displaystyle p = 22.5$$p$$=22.5$
   5. Justify: ${R}$ is continuous on ${[0, 45]}$, so the candidates test compares the critical point with the endpoints (revenue in dollars): $\displaystyle R(0) = 0$$R(0)$$=0$$\displaystyle R(45) = 0$$R(45)$$=0$$\displaystyle R(22.5) = 22.5 \times 45$$R(22.5)$$=22.5\times 45$$\displaystyle = 1{,}012.5$$=1,012.5$ The largest is at ${p = 22.5}$, so that is the absolute maximum. (Or: ${R''(p) = -4}$ is negative, so 22.5 is a relative maximum by the second derivative test; as the only critical point on ${[0, 45]}$, it is the absolute maximum.)
   6. In context: charge $22.50 a shirt. The band sells 45 and brings in $1,012.50 a night.

4. Challenge 3

   No calculator · CED 5.10, 5.11

   A car wash charges $10 and washes 120 cars on a Saturday; each 50¢ increase loses 4 cars. The number of cars is a linear function of the price, and the price can be any amount to the cent. The first steps are done for you: with ${p}$ in dollars, $1 more loses 8 cars, so for ${0 \le p \le 25}$ cars ${q}$ and revenue ${R}$ are $\displaystyle q(p) = 120 - 8(p - 10)$$q(p)$$=120-8(p-10)$$\displaystyle = 200 - 8p$$=200-8p$multiply out$\displaystyle R(p) = p(200 - 8p)$$R(p)$$=p(200-8p)$$\displaystyle = 200p - 8p^2$$=200p-8p^2$multiply out What price maximizes Saturday revenue?

   Answer: $12.50

   1. (Done for you.) ${q(p) = 200 - 8p}$ cars and ${R(p) = 200p - 8p^2}$ dollars, for ${0 \le p \le 25}$.
   2. Differentiate and set to 0: $\displaystyle R'(p) = 200 - 16p$$R^{\prime }(p)$$=200-16p$$\displaystyle 200 - 16p = 0$$200-16p$$=0$$\displaystyle p = 12.5$$p$$=12.5$
   3. ${R''(p) = -16}$ is negative, so ${p = 12.5}$ is a relative maximum by the second derivative test; as the only critical point, it's the absolute maximum on ${[0, 25]}$.
   4. Charge $12.50: ${200 - 8(12.5) = 100}$ cars and $1,250.

   Common wrong answers

   • $10.00: That's today's price. Find where R′(p) = 0.
   • $1,250.00: That's the most revenue. The question asks for the price.
   • $25.00: At $25 nobody comes: revenue is 0. Solve R′(p) = 0.

5. Challenge 4

   No calculator · CED 5.5, 5.11

   Hockey again: 900 fans want tickets at $8, and every 10¢ increase loses 6 fans. The number of fans is a linear function of the price, and the price can be any amount to the cent. Now each fan also spends $3 on concessions, and the rink holds 750: at most 750 tickets. What ticket price maximizes total revenue per game, tickets plus concessions?

   Answer: $10.50

   1. While everyone fits, total revenue in dollars is tickets plus $3 a fan in concessions: $\displaystyle T(p) = p(1{,}380 - 60p) + 3(1{,}380 - 60p)$$T(p)$$=p(1,380-60p)+3(1,380-60p)$$\displaystyle = -60p^2 + 1{,}200p + 4{,}140$$=-60p^2+1,200p+4,140$multiply out
   2. ${T'(p) = -120p + 1{,}200}$ is 0 at ${p = 10}$. But at $10, ${1{,}380 - 60(10) = 780}$ fans want tickets; only 750 fit.
   3. Demand is at most 750 when ${1{,}380 - 60p \le 750}$, so when ${p \ge 10.5}$. Below $10.50 the rink sells out and the team gets ${(p + 3) \cdot 750}$, which grows with the price. So the best price is at least $10.50.
   4. On ${[10.5, 23]}$, ${T'(p) = 1{,}200 - 120p}$ is negative: ${T}$ decreases, with no critical point inside. By the candidates test the maximum is at an endpoint (total revenue in dollars): $\displaystyle T(10.5) = 13.5 \times 750$$T(10.5)$$=13.5\times 750$$\displaystyle = 10{,}125$$=10,125$$\displaystyle T(23) = 0$$T(23)$$=0$
   5. Charge $10.50: a 750-fan sellout, and tickets plus concessions bring in $10,125.

   Common wrong answers

   • $10.00: 780 fans won't fit in a 750-seat rink.
   • $11.50: That's the best price for tickets alone. Add the $3 a fan in concessions.
   • $10,125.00: That's the most total revenue. The question asks for the price.

6. Challenge 5

   No calculator · CED 5.10, 5.11

   Phone cases: each costs you $5, and there are no other costs. At $15 you sell 60 a month; each $1 increase loses 3 sales a month. Monthly sales are a linear function of the price, and the price can be any amount to the cent. What price maximizes monthly profit?

   Answer: $20.00

   1. Let ${p}$ be the price in dollars. Cases sold a month: $\displaystyle q(p) = 60 - 3(p - 15)$$q(p)$$=60-3(p-15)$$\displaystyle = 105 - 3p$$=105-3p$multiply out
   2. Each case earns ${p - 5}$ dollars. Below $5 each sale loses money; above $35 you sell none. So for ${5 \le p \le 35}$, monthly profit in dollars is $\displaystyle P(p) = (p - 5)(105 - 3p)$$P(p)$$=(p-5)(105-3p)$$\displaystyle = -3p^2 + 120p - 525$$=-3p^2+120p-525$multiply out
   3. ${P'(p) = -6p + 120}$ is 0 at ${p = 20}$.
   4. Candidates test on ${[5, 35]}$: ${P(5) = 0}$, ${P(35) = 0}$, and ${P(20) = 15 \times 45 = 675}$ dollars. The maximum is at ${\boxed{p = 20}}$: charge $20.

   Common wrong answers

   • $17.50: $17.50 maximizes revenue, not profit. Subtract the $5 cost first.
   • $15.00: That's today's price. Find where P′(p) = 0.
   • $675.00: That's the most profit. The question asks for the price.

7. Challenge 6

   No calculator · CED 5.11

   At the best price, what is your monthly profit from phone cases?

   Answer: $675.00

   1. At $20 you sell ${105 - 3(20) = 45}$ cases a month, each earning ${20 - 5 = 15}$ dollars.
   2. The maximum monthly profit is ${45 \times 15 = \boxed{675}}$ dollars: $675 a month.

   Common wrong answers

   • $900.00: That's revenue: 45 cases at $20. Subtract the $5 cost per case.
   • $600.00: That's profit at today's $15. The best price earns more.
   • 45: That's cases sold. Multiply by the profit on each.

8. Fluency 1

   No calculator · CED 5.2

   At what value of ${x}$ does ${f(x) = x^2 + \dfrac{16}{x}}$, for ${x > 0}$, have a critical point?

   Answer: 2

   1. ${f'(x) = 2x - \dfrac{16}{x^2}}$, defined for every ${x > 0}$.
   2. Set ${f'(x) = 0}$: $\displaystyle 2x - \dfrac{16}{x^2} = 0$$2x-\frac{16}{x^2}$$=0$$\displaystyle 2x = \dfrac{16}{x^2}$$2x$$=\frac{16}{x^2}$$\displaystyle x^3 = 8$$x^3$$=8$multiply by ${x^2}$ and divide by ${2}$
   3. The cube root of 8 is 2: the critical point is at ${\boxed{x = 2}}$.

   Common wrong answers

   • 8: x³ = 8, so take the cube root.
   • 12: That's f(2). The question asks for the x where f′(x) = 0.

9. Fluency 2

   No calculator · CED 5.4

   The derivative of ${f}$ is ${f'(x) = (x - 4)(x + 1)^2}$. At what value of ${x}$ does ${f}$ have a relative minimum?

   Answer: 4

   1. ${f'(x) = 0}$ at ${x = 4}$ and ${x = -1}$.
   2. ${(x + 1)^2}$ is never negative, so ${f'}$ has the sign of ${x - 4}$: negative for ${x < 4}$ (both sides of ${-1}$), positive for ${x > 4}$.
   3. At ${x = -1}$, ${f'}$ doesn't change sign: no extremum. At ${x = 4}$ it goes negative to positive: a relative minimum at ${\boxed{x = 4}}$.

   Common wrong answers

   • -1: (x + 1)² is never negative, so f′ doesn't change sign at x = −1. No extremum.

10. Fluency 3

    No calculator · CED 5.4

    At what value of ${x}$ does ${f(x) = x^3 - 3x^2 - 9x + 4}$ have a relative maximum?

    Answer: -1

    1. Differentiate and factor: $\displaystyle f'(x) = 3x^2 - 6x - 9$$f^{\prime }(x)$$=3x^2-6x-9$$\displaystyle = 3(x - 3)(x + 1)$$=3(x-3)(x+1)$factor So ${f'(x) = 0}$ at ${x = -1}$ and ${x = 3}$.
    2. ${f'}$ is positive for ${x < -1}$, negative between ${-1}$ and ${3}$, and positive for ${x > 3}$.
    3. ${f'}$ goes positive to negative at ${x = -1}$: a relative maximum at ${\boxed{x = -1}}$. (At ${x = 3}$, a relative minimum.)

    Common wrong answers

    • 3: That's the relative minimum: f′ goes negative to positive at x = 3.
    • 9: That's f(−1), the maximum value. The question asks where.

11. Fluency 4

    No calculator · CED 5.7

    ${f(x) = x^3 - 12x + 1}$ has critical points at ${x = -2}$ and ${x = 2}$. Use the second derivative test to find the relative minimum value of ${f}$.

    Answer: -15

    1. ${f''(x) = 6x}$, so ${f''(-2) = -12}$ and ${f''(2) = 12}$.
    2. At ${x = 2}$, ${f'(2) = 0}$ and ${f''(2) = 12}$ is positive: a relative minimum by the second derivative test. At ${x = -2}$, ${f''(-2) = -12}$ is negative: a relative maximum.
    3. The relative minimum value is ${f(2) = 8 - 24 + 1 = \boxed{-15}}$.

    Common wrong answers

    • 2: The relative minimum is at x = 2. The question asks for f(2).
    • 17: That's f(−2), the relative maximum value. f″(−2) < 0 there.

12. Fluency 5

    No calculator · CED 5.5

    Find the absolute maximum value of ${f(x) = 12x - x^3}$ on ${[0, 3]}$.

    Answer: 16

    1. ${f'(x) = 12 - 3x^2}$, which is 0 at ${x = \pm 2}$. Only ${x = 2}$ is in ${[0, 3]}$.
    2. Candidates: ${f(0) = 0}$, ${f(2) = 24 - 8 = 16}$, and ${f(3) = 36 - 27 = 9}$.
    3. The absolute maximum value is 16, at ${x = 2}$.

    Common wrong answers

    • 2: The maximum is at x = 2. The question asks for f(2).
    • 9: That's f(3), an endpoint. Check the critical point too.

13. Fluency 6

    No calculator · CED 5.5

    Find the absolute minimum value of ${g(x) = 2x^3 - 9x^2 + 12x}$ on ${[0, 3]}$.

    Answer: 0

    1. Differentiate and factor: $\displaystyle g'(x) = 6x^2 - 18x + 12$$g^{\prime }(x)$$=6x^2-18x+12$$\displaystyle = 6(x - 1)(x - 2)$$=6(x-1)(x-2)$factor So ${g'(x) = 0}$ at ${x = 1}$ and ${x = 2}$.
    2. Candidates: ${g(0) = 0}$, ${g(1) = 5}$, ${g(2) = 4}$, and ${g(3) = 9}$.
    3. The relative minimum at ${x = 2}$ isn't lowest; the endpoint is. The absolute minimum value is 0, at ${x = 0}$.

    Common wrong answers

    • 4: That's the relative minimum, g(2). The endpoint g(0) is lower.
    • 2: That's where the relative minimum is. Compare g at every candidate.

14. Review 1

    From Is the Ball Ever Stopped?, Challenge 1 · No calculator · CED 1.1, 2.1

    A pop-up's height after ${t}$ seconds is ${h(t) = {6 + 80t - 16t^2}}$ feet. Find its average velocity from ${t = 1}$ to ${t = 2}$ seconds.

    Answer: 32

    1. Average velocity: ${\dfrac{h(2) - h(1)}{2 - 1}}$.
    2. Heights, in feet: $\displaystyle h(1) = 6 + 80 - 16$$h(1)$$=6+80-16$$\displaystyle = 70$$=70$$\displaystyle h(2) = 6 + 160 - 64$$h(2)$$=6+160-64$$\displaystyle = 102$$=102$
    3. ${\dfrac{102 - 70}{2 - 1} = \boxed{32}}$: the ball rises 32 feet in that second, an average velocity of 32 ft/s.

    Common wrong answers

    • 86: That's the average height, not change in height over change in time.
    • 48: That's the average from t = 0 to t = 2. Start at t = 1.
    • -32: Same order on top and bottom: h(2) − h(1) over 2 − 1.
    • 102: That's the height at t = 2, not a velocity.

15. Review 2

    From Is the Ball Ever Stopped?, Fluency 5 · No calculator · CED 2.2

    Use the limit definition to find ${f'(x)}$ for ${f(x) = 5x^2 + 1}$. Then evaluate ${f'(-2)}$.

    Answer: -20

    1. $\displaystyle f(x + h) = 5(x + h)^2 + 1$$f(x+h)$$=5(x+h{)}^2+1$$\displaystyle = 5x^2 + 10xh + 5h^2 + 1$$=5x^2+10xh+5h^2+1$multiply out
    2. Subtract ${f(x) = 5x^2 + 1}$, then divide by ${h}$: $\displaystyle f(x + h) - f(x) = 10xh + 5h^2$$f(x+h)-f(x)$$=10xh+5h^2$$\displaystyle \frac{f(x + h) - f(x)}{h} = 10x + 5h$$\frac{f(x+h)-f(x)}{h}$$=10x+5h$${h \ne 0}$
    3. As ${h \to 0}$, this goes to ${f'(x) = 10x}$, so ${f'(-2) = \boxed{-20}}$.

    Common wrong answers

    • 21: That's f(−2). Find f′(x), then substitute x = −2.
    • 20: Check the sign: f′(x) = 10x, and x is −2.

16. Exit ticket

    No calculator · CED 5.10, 5.11

    Team-store hoodies cost $12 each to make, with no other costs. At $30 the store sells 80 a week; each $1 increase loses 2 sales a week. Weekly sales are a linear function of the price, and the price can be any amount to the cent. What price maximizes weekly profit?

    Answer: $41.00

    1. Let ${p}$ be the price in dollars. Hoodies sold a week: $\displaystyle q(p) = 80 - 2(p - 30)$$q(p)$$=80-2(p-30)$$\displaystyle = 140 - 2p$$=140-2p$multiply out
    2. Each hoodie earns ${p - 12}$ dollars, so for ${12 \le p \le 70}$ weekly profit in dollars is $\displaystyle P(p) = (p - 12)(140 - 2p)$$P(p)$$=(p-12)(140-2p)$$\displaystyle = -2p^2 + 164p - 1{,}680$$=-2p^2+164p-1,680$multiply out
    3. ${P'(p) = -4p + 164}$ is 0 at ${p = 41}$, and ${P''(p) = -4}$ is negative: a relative maximum by the second derivative test. As the only critical point on ${[12, 70]}$, it is the absolute maximum.
    4. Charge $41: 58 hoodies a week, and ${29 \times 58 = 1{,}682}$ dollars of profit a week.

    Common wrong answers

    • $35.00: $35 maximizes revenue, not profit. Subtract the $12 cost first.
    • $30.00: That's today's price. Find where P′(p) = 0.