Maximizing Revenue: Price & Attendance Trade-off
An amusement park charges admission price (P) and averages 2000 tickets (T) sold per day. A survey shows that, for each increase in the admission cost, 100 fewer people would visit the park. Write an equation to express the revenue (R), in terms of a price increase of x dollars and then determine the admission price that gives the maximum revenue.
A note on the setup. The problem names the current price P but does not give it a numerical value, and "each increase" is read as each $1 increase. So the honest answer is a formula in P. We solve it in general and then plug in a concrete price ($10) so you can see actual numbers. If your textbook or worksheet supplies a value for P, substitute it into the final formulas.
Already Got the Answer? Check It Here
Setup:R = (P + x)(2000 − 100x) = −100x² + (2000 − 100P)x + 2000P
Maximum at:x = 10 − P/2, so the best admission price is P + x = 10 + P/2 = (P + 20)/2 dollars.
Example (P = $10): x = 5, price $15, 1500 tickets, R = $22,500.
One-line check: at P = $10, a $14 ticket gives 14 × 1600 = $22,400 and a $16 ticket gives 16 × 1400 = $22,400, both below $22,500.
Skills This Problem Builds
- Revenue as a product of two changing quantities. When a price goes up, the quantity sold goes down. The product of two linear expressions is a quadratic, and that is where the "peak" comes from.
- Choosing a variable for the change, not the total. Defining
xas the increase makes the quantity formula simple, but thenxis not the answer to "what price?" - Three ways to locate a parabola's maximum: the vertex formula, the midpoint of the zeros, and step-by-step marginal comparison.
- Working with a symbolic parameter. Because
Pis unspecified, you get to see how the answer depends on it, including the case where raising the price is the wrong move. - Judging whether a model's answer is allowed. A vertex can fall outside the realistic domain, so you check it against the story.
Solution: Method 1 — Building the Revenue Function and Finding Its Vertex
Revenue is always price × quantity sold. The whole problem is about writing both factors in terms of one variable, the price increase.
Step 1 — Define the variable and the two factors
Let x be the number of dollars the admission price goes up. The new price is the old price plus the increase, and every dollar of increase costs the park 100 visitors:
tickets = 2000 − 100x
Step 2 — Multiply to get the revenue equation
This answers the first half of the question:
Expanding with the distributive property:
R = −100x² + (2000 − 100P)x + 2000P
This is a quadratic in x with a = −100, b = 2000 − 100P, and c = 2000P. Since a is negative, the parabola opens downward and has a highest point.
Step 3 — Find the vertex
A downward parabola peaks at x = −b/(2a):
x = (2000 − 100P) / 200
x = 10 − P/2
Step 4 — Turn x back into a price
The question asks for the admission price, not the increase. Add the increase to the starting price:
Plugging this into the equation gives the maximum revenue. Tickets sold at this price are 2000 − 100(10 − P/2) = 1000 + 50P, so:
Step 5 — Try it with real numbers
Suppose the current price is P = $10. Then:
x = −1000 / (2 · (−100)) = 5
price = 10 + 5 = $15, tickets = 2000 − 500 = 1500
R = 15 × 1500 = $22,500
Domain matters. The model only makes sense for 0 ≤ x ≤ 20: no negative increases (the problem says "increase") and no negative attendance (tickets reach 0 at x = 20). The vertex x = 10 − P/2 is a legitimate answer only when it lands in that interval, which for x ≥ 0 means P ≤ 20.
Solution: Method 2 — Symmetry of the Parabola in Terms of Price
You can find the peak without the vertex formula. A parabola is perfectly symmetric, so its maximum sits exactly halfway between its two zeros. The trick is to work with the price p directly instead of the increase.
Step 1 — Express attendance in terms of the price p
The increase is x = p − P, so tickets are 2000 − 100(p − P) = 100(20 + P − p). Revenue is then:
Step 2 — Read off the zeros
Revenue is zero when p = 0 (a free ticket earns nothing) and when p = 20 + P (the price is so high that nobody comes).
Step 3 — Take the midpoint
This matches Method 1 exactly. With P = $10 the zeros are $0 and $30, the peak is at $15, and R = 100 · 15 · 15 = $22,500.
The result is a nice rule of thumb: the revenue-maximizing price is half of the "nobody comes" price. Any linear demand model has this feature.
Solution: Method 3 — Marginal Reasoning with a Table
No algebra tricks here. Take P = $10, raise the price one dollar at a time, and watch what happens to revenue.
| x (increase) | Price | Tickets | Revenue | Change from previous |
|---|---|---|---|---|
| 0 | $10 | 2000 | $20,000 | — |
| 1 | $11 | 1900 | $20,900 | +900 |
| 2 | $12 | 1800 | $21,600 | +700 |
| 3 | $13 | 1700 | $22,100 | +500 |
| 4 | $14 | 1600 | $22,400 | +300 |
| 5 | $15 | 1500 | $22,500 | +100 |
| 6 | $16 | 1400 | $22,400 | −100 |
| 7 | $17 | 1300 | $22,100 | −300 |
| 8 | $18 | 1200 | $21,600 | −500 |
Look at the last column. The gains shrink by exactly 200 each step (900, 700, 500, 300, 100) and then turn into losses. That constant drop is the fingerprint of a quadratic: its "second difference" is 2a = −200. Revenue peaks at the last step where the change is still positive.
In general, going from x to x + 1 changes revenue by 1900 − 100P − 200x. Setting this to zero gives x = 9.5 − P/2, which is the vertex shifted by half a step, as expected when you move in whole-dollar increments. For P = 10 it means the increases at x = 5 still gain and the one after it loses.
The Answer
Revenue equation (in terms of the increase x):
Revenue-maximizing price:
Maximum revenue:R = 25(P + 20)² dollars per day.
With a current price of $10: raise it by $5 to $15, sell 1500 tickets, and earn $22,500 per day, up from $20,000.
A price increase helps only when P < 20. At $20 the park is already at its best price, and above $20 the best move is to lower the price.
Verification
Check three things with P = $10: the equation, the vertex, and the neighbors.
- Equation at x = 0:
(10 + 0)(2000 − 0) = 20,000, which equals the starting revenue of 10 × 2000, as it must. - Equation at x = 5:
(10 + 5)(2000 − 500) = 15 × 1500 = 22,500. The expanded form agrees:−100(25) + 1000(5) + 20000 = −2500 + 5000 + 20000 = 22,500. - Neighbors are lower:
x = 4gives 14 × 1600 = 22,400 andx = 6gives 16 × 1400 = 22,400. Both are less than 22,500, so x = 5 is the peak.
The general formula passes a second test: 25(P + 20)² at P = 10 is 25 × 900 = 22,500, and at P = 20 it is 25 × 1600 = 40,000 = 20 × 2000, which is just today's revenue, as expected because x = 0 is optimal there.
Reality Check
With P = $10, the price rises 50% (from $10 to $15) while attendance falls only 25% (from 2000 to 1500). Since 1.50 × 0.75 = 1.125, revenue rises 12.5%, from $20,000 to $22,500. That is a modest gain, which is reasonable. If a quadratic peak promised a 300% gain from a small price change, something would be wrong.
Also notice that the peak is flat: a $1 error on either side costs only $100 (0.4%). Real pricing decisions are forgiving near the optimum, so being roughly right matters more than being exactly right.
What Trips People Up
✗ R = P(2000 − 100x)
This keeps the price fixed at P while the attendance drops, so it forgets that the price is exactly what is rising. Both factors of revenue must depend on x: price is P + x, not P.
✗ tickets = 2000 − 100(P + x)
The 100-visitor loss applies per dollar of increase, not per dollar of the whole price. Attendance is 2000 at the current price, so the starting point is x = 0, not x = −P.
✗ "The answer is x = 5 dollars."
That is the increase. The question asks for the admission price, which is P + x, here $15. Whenever you define a variable as a change, finish by converting back to the quantity asked for.
✗ Accepting x = −10 when P = $40
With a $40 ticket, the vertex formula gives x = 10 − 20 = −10, a $10 price decrease. The algebra is fine, but the problem is about increases, and the model also assumes attendance responds in the same linear way to price cuts. Check the domain before reporting a vertex.
The General Formula and Its Family
Any "price up, quantity down" problem with a linear demand has the same skeleton. If the starting price is P, the starting quantity is Q, and each $1 increase loses m customers:
x* = (Q − m·P) / (2m)
p* = (Q/m + P) / 2 (half of the "nobody comes" price Q/m + P)
Here Q = 2000 and m = 100, so Q/m = 20 and p* = (20 + P)/2. This is the same parabola you meet in fence-and-area problems, projectile height, and profit models. Whenever two linear quantities are multiplied and one rises as the other falls, a downward parabola appears.
Limitations: the linear-demand assumption is a local approximation. Real demand curves bend, and the formula can't be trusted far from the observed prices. In calculus terms, the maximum is where dR/dx = 2000 − 100P − 200x = 0, which gives the same x.
How to Spot This Problem Type
- "For each $1 increase, N fewer…" — a linear trade-off between price and quantity.
- "Maximum revenue" or "maximum profit" — a quadratic with a downward opening.
- "In terms of a price increase of x" — the problem is telling you which variable to define.
- Same structure in disguise: "A farmer has 30 trees per acre and each extra tree reduces yield by 2 apples per tree; how many trees per acre maximize yield?" That is (30 + x)(Y − 2x) again, with trees in place of price.
Beyond the Textbook
- Ticket and subscription pricing: theaters, concerts, and software companies run price experiments to estimate exactly this kind of "customers lost per dollar" slope.
- Economics: the point where the marginal gain from the higher price equals the marginal loss from lost customers is the textbook condition for revenue maximization, and it is Method 3 with a smaller step.
- Hotels and airlines: yield management adjusts prices up and down to sit near the top of a revenue parabola as demand shifts.
If You Want to Go Deeper
Revenue is not profit. Once the park has costs, the optimal price changes: a ticket price that maximizes revenue is generally too low to maximize profit, because every lost visitor also saves the park some cost. The fourth What-If below explores exactly this. Beyond that, you can replace the straight-line demand with a curved one and use calculus (or elasticity, the percent change in quantity per percent change in price) to find the peak.
Extend Your Thinking
Work each problem on paper first, then open the solution to compare.
A small park charges $4 admission and averages 2000 tickets per day. This time, each $1 increase in price loses 200 visitors. Write the revenue equation in terms of the increase x, and find the admission price that maximizes revenue and the maximum revenue.
Price is 4 + x. Tickets are 2000 − 200x.
R = (4 + x)(2000 − 200x) = −200x² + 1200x + 8000.
x = −1200 / (2 · (−200)) = 3.
Price: 4 + 3 = $7. Tickets: 2000 − 600 = 1400. Revenue: 7 × 1400 = 9800. The best price is $7, with a maximum revenue of $9,800 per day.
At x = 2: 6 × 1600 = 9,600. At x = 4: 8 × 1200 = 9,600. Both are below 9,800, so x = 3 is the peak. Notice that doubling the sensitivity pulled the optimum price down compared with the 100-visitor case, where the zeros would sit at $0 and $24.
A consultant tells the park (2000 tickets per day, 100 fewer visitors per $1 increase) that the revenue-maximizing admission price is $18. What is the park's current price P, and what is the maximum daily revenue?
For this park, the best price is 10 + P/2. This comes from R = (P + x)(2000 − 100x) with vertex x = 10 − P/2.
10 + P/2 = 18, so P/2 = 8 and P = 16.
x = 10 − 16/2 = 2, so the current $16 price is raised by $2 to $18.
Tickets: 2000 − 200 = 1800. Revenue: 18 × 1800 = 32,400. The current price is $16 and the maximum revenue is $32,400 per day.
The formula 25(P + 20)² = 25 × 36² = 25 × 1296 = 32,400 matches. Neighbors: x = 1 gives 17 × 1900 = 32,300 and x = 3 gives 19 × 1700 = 32,300, both lower.
The park charges $10 admission, averages 2000 tickets per day, and loses 100 visitors for each $1 increase. Its fixed daily costs are $22,400. For which admission prices does the park earn revenue greater than its costs?
R = (10 + x)(2000 − 100x) = −100x² + 1000x + 20000.
We need R > 22,400, so −100x² + 1000x + 20000 > 22400, which simplifies to −100x² + 1000x − 2400 > 0.
Divide by −100 and flip the inequality: x² − 10x + 24 < 0, which factors as (x − 4)(x − 6) < 0.
A product of these factors is negative only between the roots, so 4 < x < 6. In prices, 10 + x lies between 14 and 16. The park covers its costs when the admission price is strictly between $14 and $16.
At $15 (x = 5): 15 × 1500 = 22,500 > 22,400. At $14 and $16 the revenue is exactly 22,400 (break-even). At $13: 13 × 1700 = 22,100 and at $17: 17 × 1300 = 22,100, both below cost. The window is centered on the $15 peak, as symmetry predicts.
The park charges $10 admission and averages 2000 tickets per day, losing 100 visitors for each $1 increase. Each visitor costs the park $6 to serve (staffing, cleaning, ride upkeep). Write a daily profit equation in terms of the increase x, find the price that maximizes profit, and compare it with the revenue-maximizing price of $15.
A visitor pays 10 + x and costs $6, so each one contributes (10 + x) − 6 = 4 + x dollars.
Profit = (4 + x)(2000 − 100x) = −100x² + 1600x + 8000.
x = −1600 / (2 · (−100)) = 8. (Or use the zeros: x = −4 and x = 20, midpoint 8.)
Price: 10 + 8 = $18. Visitors: 2000 − 800 = 1200. Profit: 12 × 1200 = 14,400. The profit-maximizing price is $18, which is higher than the $15 that maximizes revenue.
At $15 the profit is only 9 × 1500 = 13,500, which is less than 14,400. Neighbors of $18: x = 7 gives 11 × 1300 = 14,300 and x = 9 gives 13 × 1100 = 14,300, both lower. The park earns more profit by serving fewer people at a higher price, because each lost visitor also saves $6 in costs.
Frequently Asked Questions
2026-05-20