Pricing a One-Off Concert
A promoter is booking a 1,200-capacity hall for a one-off concert. At a ticket price of €25 she expects to sell all 1,200 seats. Research on comparable shows suggests that for every €1 added to the price, attendance falls by 40; for every €1 cut, it rises by 40, up to the hall's capacity. The hall costs €18,000 to hire, and each attendee costs a further €3 in staffing and ticketing fees, whether or not the show sells out. Prices must be quoted in whole euros. Before signing, the promoter has told the venue she expects the show to clear at least €9,500 in profit.
1. Which whole-euro ticket price maximises revenue, and what is that maximum?
Read the worked answer
Let x be the change in price in euros. Price = 25 + x; attendance = 1,200 − 40x.
Revenue R(x) = (25 + x)(1,200 − 40x) = 30,000 + 1,200x − 1,000x − 40x² = 30,000 + 200x − 40x².
This is a downward parabola; its peak is at x = −200 / (2 × −40) = 2.5.
But prices are whole euros, so x must be an integer. Check the neighbours:
- x = 2: €27 × 1,120 = €30,240
- x = 3: €28 × 1,080 = €30,240
- x = 2.5 would give €30,250, but is not allowed.
Answer: €27 or €28 both maximise revenue at €30,240. They are tied because the parabola is symmetric about x = 2.5. Any answer quoting €30,250 has ignored the whole-euro constraint.
2. Which whole-euro price maximises profit, and how much is that profit? Explain why it is not the revenue-maximising price.
Read the worked answer
Attendance costs €3 per head, so profit is not maximised where revenue is:
P(x) = (25 + x)(1,200 − 40x) − 18,000 − 3(1,200 − 40x) = (1,200 − 40x)[(25 + x) − 3] − 18,000 = (1,200 − 40x)(22 + x) − 18,000 = 26,400 + 1,200x − 880x − 40x² − 18,000 = 8,400 + 320x − 40x²
Peak at x = −320 / (2 × −40) = 4, an integer. So:
- Price: 25 + 4 = €29
- Attendance: 1,200 − 160 = 1,040
- Revenue: 29 × 1,040 = €30,160
- Costs: 18,000 + 3 × 1,040 = €21,120
- Profit: €9,040
Why not €27–€28: below the profit optimum, each €1 price increase adds 200 to revenue but saves 3 × 40 = €120 of attendee cost, a net gain of €80 up to x = 4. Past x = 4 the revenue lost per €1 (which grows as attendance shrinks) outweighs the cost saved. Profit peaks where revenue has already started to fall - the standard result when serving a customer is itself costly.
3. Can the promoter deliver the €9,500 profit she has promised the venue? Justify your answer with figures, and say what she would have to change.
Read the worked answer
No. The maximum is €9,040, at €29. Check the neighbourhood for completeness:
- x = 3 (€28, 1,080 sold): 30,240 − 18,000 − 3,240 = €9,000
- x = 5 (€30, 1,000 sold): 30,000 − 18,000 − 3,000 = €9,000
So €9,040 is the true ceiling, and the shortfall against the promise is €460.
The reason is structural, not a matter of trying harder: P(x) = 8,400 + 320x − 40x² has a maximum of €9,040 at any x, because the €18,000 hire is fixed and unrecoverable by pricing. No ticket price - and no combination of price and discount - reaches €9,500.
What must change:
- Cut the fixed cost by at least €460 through negotiation with the venue; or
- Reduce the €3 per-head cost to €2.54 or less (profit rises by €1 × attendance, ≈ €1,040); or
- Add revenue that does not vary with attendance (sponsorship, bar commission) of €460 or more.
A good answer states the shortfall as a precise number and identifies it as a fixed-cost problem rather than a pricing problem.