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Best Price · Unit 5

What ticket price makes the most money?

Raise the price and you lose customers; lower it and each sale earns less. Critical points, the first and second derivative tests, and optimization.

About 45 min · 13 problemsStart →

Checks grade only your final answer. The write-up unlocks when you're right or after two tries; hints are always open.

Two prices to set

The hockey tickets. At $8 a ticket, 900 fans want to come; each 10¢ more loses 6.

The phone cases. Each costs you $5; at $15 you sell 60 a month.

Both: a raise earns more per sale but sells fewer, so the goal rises, then falls. Optimization; the tool is the derivative.

Predict first

Raise or lower?

Friend, Teammate, Coach, You

  1. Friend
    raise it. more money per ticket

Price the tickets

No calculus yet: R=p⋅n{R = p \cdot n}, price times fans. At $8, 900 fans and $7,200; at $9, 840 fans pay $9 each and bring in $7,560. Where does raising stop helping?

Challenge 1

No calculator
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?

Dollars and cents, like 4.25 or $1,200.

Challenge 2

No calculator
Same team. What is the most ticket revenue per game?

Dollars and cents, like 4.25 or $1,200.

Slide the price

Revenue from $8 to $23, where nobody comes. The tangent's slope is R′(p){\teal{R'(p)}}, dollars of revenue per dollar of price.

Ticket revenue at every priceRevenue model: R(p)=p(1,380−60p){R(p) = p(1{,}380 - 60p)}. A synthetic model.

Drag across the graph, or use the slider, to move the point.

Ticket price

Tangent at $8.00: revenue $7,200.00, slope 420 dollars of revenue per dollar of price.

Try these:

  1. At $8 the slope is 420, but the real gain from $8 to $9 is only $360. Why less?
  2. Drag until the tangent is flat. Price and revenue there? Check Challenges 1 and 2.
  3. At $15 the slope is −420. What should the team do?
  4. $11 and $12 both bring in $7,920. Where is the flat tangent, and why?
Your prediction

Answer: raise it. At $8 the tangent's slope is 420, so revenue climbs as the price goes up.

About the model

The team, fans, and numbers are made up. Real demand isn't a straight line, but near today's price a line is often a good model.

The top of the hill

Inside a smooth graph, a peak or valley has a flat tangent: look first where the derivative is 0.

  • Critical point. An x{x} in the domain of f{f} where f′(x){f'(x)} is 0 or doesn't exist. Every relative extremum is at one, but not the reverse: x3{x^3} is flat at 0 and keeps climbing.
  • First derivative test. At a critical point c{c} where f{f} is continuous: f′{f'} positive to negative, relative maximum; negative to positive, relative minimum; no sign change, neither.
  • Second derivative test. f′(c)=0{f'(c) = 0} and f′′(c)<0{f''(c) < 0}: relative maximum. f′′(c)>0{f''(c) > 0}: relative minimum. f′′(c)=0{f''(c) = 0}: no conclusion; use the first derivative test.
  • Candidates test. A continuous f{f} on [a,b]{[a, b]} has an absolute maximum and minimum (Extreme Value Theorem), at critical points or endpoints. Compare f{f} at each.
  • Only one critical point. On an interval, a continuous function's only critical point, if a relative maximum, is the absolute maximum. Same for a minimum.

Optimization, every time:

  1. Variable: name it, with units.
  2. Function: the quantity to optimize, in that one variable.
  3. Domain: what the story allows.
  4. Critical points inside the domain.
  5. Justify: a derivative test plus the only-critical-point rule, or compare every candidate, endpoints included.
  6. Answer in context, with units.

Worked example, then your turn

The six steps, worked:

Worked example

No calculator
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?
  1. Variable and units: p{p} is the price in dollars. Sales drop 2 per dollar above $20, so shirts sold are q(p)=50−2(p−20)\displaystyle q(p) = 50 - 2(p - 20)=90−2p\displaystyle = 90 - 2pmultiply out
  2. Function: revenue in dollars is price times shirts sold: R(p)=p⏟price⋅(90−2p)⏟shirts sold\displaystyle R(p) = \underbrace{p}_{\text{price}} \cdot \underbrace{(90 - 2p)}_{\text{shirts sold}}=90p−2p2\displaystyle = 90p - 2p^2multiply out
  3. Domain: price and sales can't be negative, so 0≤p≤45{0 \le p \le 45}.
  4. Critical points: the tangent's slope R′(p){\teal{R'(p)}} is 0 where the tangent is flat: R′(p)=90−4p\displaystyle \teal{R'(p)} = 90 - 4pdifferentiate90−4p=0\displaystyle 90 - 4p = 0set the slope to 0p=22.5\displaystyle p = 22.5
  5. Justify: R{R} is continuous on [0,45]{[0, 45]}, so the candidates test compares the critical point with the endpoints (revenue in dollars): R(0)=0\displaystyle R(0) = 0R(45)=0\displaystyle R(45) = 0R(22.5)=22.5×45\displaystyle R(22.5) = 22.5 \times 45=1,012.5\displaystyle = 1{,}012.5 The largest is at p=22.5{p = 22.5}, so that is the absolute maximum. (Or: R′′(p)=−4{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]{[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.

Setup done; you find the price.

Challenge 3

No calculator
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{p} in dollars, $1 more loses 8 cars, so for 0≤p≤25{0 \le p \le 25} cars q{q} and revenue R{R} are q(p)=120−8(p−10)\displaystyle q(p) = 120 - 8(p - 10)=200−8p\displaystyle = 200 - 8pmultiply outR(p)=p(200−8p)\displaystyle R(p) = p(200 - 8p)=200p−8p2\displaystyle = 200p - 8p^2multiply out What price maximizes Saturday revenue?

Dollars and cents, like 4.25 or $1,200.

The rink has a limit

Two changes: each fan also spends $3 on concessions, and the rink holds only 750.

Challenge 4

No calculator
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?

Dollars and cents, like 4.25 or $1,200.

What the seat cap does

Friend, Teammate, Coach

  1. Friend
    with concessions the derivative is 0 at $10, so charge $10

The side business

Phone cases again: each costs you $5, so maximize profit, not revenue.

Challenge 5

No calculator
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?

Dollars and cents, like 4.25 or $1,200.

Challenge 6

No calculator
At the best price, what is your monthly profit from phone cases?

Dollars and cents, like 4.25 or $1,200.

Check both on the profit curve: drag until the tangent is flat, slope P′(p)=0{\teal{P'(p)} = 0}.

Monthly profit at every priceProfit model: P(p)=(p−5)(105−3p){P(p) = (p - 5)(105 - 3p)}. A synthetic model.

Drag across the graph, or use the slider, to move the point.

Price

Tangent at $15.00: monthly profit $600.00, slope 30 dollars of profit per dollar of price.

Your prediction

Answer: different prices. The price for the most money is $17.50. The best profit price is higher: every sale a raise loses also saves you $5.

Fluency

Six quick ones: critical points, both derivative tests, and the candidates test.

Fluency 1

No calculator
At what value of x{x} does f(x)=x2+16x{f(x) = x^2 + \dfrac{16}{x}}, for x>0{x > 0}, have a critical point?

A number or fraction, like 12, 0.25, or 3/4.

Fluency 2

No calculator
The derivative of f{f} is f′(x)=(x−4)(x+1)2{f'(x) = (x - 4)(x + 1)^2}. At what value of x{x} does f{f} have a relative minimum?

A number or fraction, like 12, 0.25, or 3/4.

Fluency 3

No calculator
At what value of x{x} does f(x)=x3−3x2−9x+4{f(x) = x^3 - 3x^2 - 9x + 4} have a relative maximum?

A number or fraction, like 12, 0.25, or 3/4.

Fluency 4

No calculator
f(x)=x3−12x+1{f(x) = x^3 - 12x + 1} has critical points at x=−2{x = -2} and x=2{x = 2}. Use the second derivative test to find the relative minimum value of f{f}.

A number or fraction, like 12, 0.25, or 3/4.

Fluency 5

No calculator
Find the absolute maximum value of f(x)=12x−x3{f(x) = 12x - x^3} on [0,3]{[0, 3]}.

A number or fraction, like 12, 0.25, or 3/4.

Fluency 6

No calculator
Find the absolute minimum value of g(x)=2x3−9x2+12x{g(x) = 2x^3 - 9x^2 + 12x} on [0,3]{[0, 3]}.

A number or fraction, like 12, 0.25, or 3/4.

Mixed review

From earlier lessons: Is the Ball Ever Stopped?

Review 1

No calculator
A pop-up's height after t{t} seconds is h(t)=6+80t−16t2{h(t) = {6 + 80t - 16t^2}} feet. Find its average velocity from t=1{t = 1} to t=2{t = 2} seconds.

A number or fraction, like 12, 0.25, or 3/4.

Review 2

No calculator
Use the limit definition to find f′(x){f'(x)} for f(x)=5x2+1{f(x) = 5x^2 + 1}. Then evaluate f′(−2){f'(-2)}.

A number or fraction, like 12, 0.25, or 3/4.

Exit ticket

Exit ticket

No calculator
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?

Dollars and cents, like 4.25 or $1,200.

Reflect

  1. Why is the derivative 0 at a best price inside the domain on a smooth graph? Why not in the rink?
  2. When is the best answer at an endpoint? An example from this lesson?
  3. Which prediction was off, and why?

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