Festival Ticket Pricing with a Venue Capacity Constraint
The student council is setting the price for tickets to the spring festival. Last year tickets sold for $8 and 420 were sold. A survey of students suggests that for each $1 increase in price, about 30 fewer tickets would be sold. A council member writes the model R(x) = (8 + x)(420 − 30x), where x is the price increase in dollars, and concludes that revenue is maximized at x = 7 'because that makes the two factors equal - $15 and 210 tickets.' The gym, the only available venue, holds 350 people and cannot be reconfigured, so no more than 350 tickets can be sold regardless of demand. The council must recommend a single whole-dollar ticket price at Tuesday's meeting, and the treasurer wants the recommendation to include expected revenue and a statement of what the model assumes.
1. Find the ticket price that maximizes revenue under the model, both before and after accounting for the 350-seat capacity. State the maximum expected revenue and the number of tickets sold at that price.
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Expand the model: R(x) = (8 + x)(420 − 30x) = 3360 + 180x − 30x².
Without the cap: the vertex is at x = −180/(2·(−30)) = 3. Price = $11, tickets = 420 − 90 = 330, revenue = 11 × 330 = $3,630.
With the 350 cap: demand 420 − 30x exceeds 350 whenever x < 7/3 ≈ 2.33. On that interval actual revenue is 350(8 + x), which rises with price, reaching 350 × $10.33 ≈ $3,616.67 at x = 7/3. For x ≥ 7/3 the quadratic applies and peaks at x = 3 with $3,630. Since $3,630 > $3,616.67, the cap does not change the optimum.
Check nearby whole-dollar prices:
- $10 (x = 2): demand 360, capped at 350 → $3,500
- $11 (x = 3): 330 tickets → $3,630
- $12 (x = 4): 300 tickets → $3,600
Recommendation: $11 per ticket, about $3,630 in revenue from 330 tickets.
2. Critique the member's x = 7 reasoning, explain what the vertex represents in this context, and describe the interval on which raising the price actually reduces revenue.
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The member has imported a rule that does not apply. For two numbers with a fixed sum, the product is largest when they are equal. Here the sum is not fixed: (8 + x) + (420 − 30x) = 428 − 29x, which shrinks as x grows. Setting the factors equal has no optimizing significance.
At x = 7 the price is $15 and demand is 210, giving $3,150 - $480 below the maximum. The vertex (3, 3630) means that a $3 increase to $11 is the point where the extra revenue per ticket is exactly offset by the revenue lost from unsold seats; below $11 the price effect dominates, above $11 the attendance effect dominates.
Revenue increases on 0 ≤ x < 3 and decreases for 3 < x ≤ 14 (demand hits zero at x = 14). With the seat cap, revenue still rises over 0 ≤ x < 7/3, but only because price rises while the house stays full - a different mechanism from the quadratic's trade-off. Note that x = 7 also lies well outside the observed data (only the $8, 420-ticket point was observed), so it is an extrapolation.
3. The treasurer asks how much confidence to place in the recommendation. Identify two assumptions built into R(x) and describe one kind of evidence that would seriously weaken the model.
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Assumption 1 - constant linear demand. The model asserts that every $1 increase costs exactly 30 tickets, from $8 all the way to $22. Real demand curves are usually steeper near the top of the range; the first $1 increase may cost almost nothing while the sixth costs far more than 30 tickets.
Assumption 2 - stated intent equals purchase. The −30 slope comes from a survey of what students say they would do, which typically overstates buying at high prices, and ignores competing events, weather, and whether the same students are surveyed.
Other assumptions: one price for all buyers, no variable costs, and a hard 350-seat cap with no standing room.
Evidence that would weaken the model: a record from a previous dance showing that a $2 increase cut attendance by 100 rather than 60; or survey data at several price points revealing curvature; or a comparable school's actual sales at $15 falling well below 210. Any of these would show the constant slope is the model's weak link, and the council should then refit or report a range of revenues rather than a single figure.