Concert Ticket Pricing: Vertex, Break-Even, and a Profit Target
The student council is setting the price of a winter concert ticket. A market survey suggests that if the price is x dollars, the profit in dollars is P(x) = -3x^2 + 240x - 1200, where 0 <= x <= 80. The council wants to maximize profit, needs to know the break-even prices, and has promised the principal a profit of at least $3000. A member proposes charging $60 per ticket because '60 is a nice round number near the middle of the range.' The faculty advisor asks the council to justify the price with algebra rather than intuition, and to state any restrictions the mathematical model places on acceptable prices.
1. Find the ticket price that maximizes profit and the maximum profit. Show the completing-the-square form and explain why the vertex gives a maximum.
Read the worked answer
Factor -3 from the first two terms:
P(x) = -3(x^2 - 80x) - 1200.
Complete the square inside the parentheses: x^2 - 80x = (x - 40)^2 - 1600. Therefore
P(x) = -3[(x - 40)^2 - 1600] - 1200 = -3(x - 40)^2 + 4800 - 1200 = -3(x - 40)^2 + 3600.
The vertex is (40, 3600). Because the coefficient -3 is negative, the parabola opens downward, so the vertex is a maximum. A price of $40 gives the maximum profit of $3600. This price lies within the stated domain 0 <= x <= 80.
2. Find the break-even prices exactly and interpret them in context. Why is the profit positive between the two break-even prices?
Read the worked answer
Set P(x) = 0:
-3x^2 + 240x - 1200 = 0.
Divide both sides by -3:
x^2 - 80x + 400 = 0.
By the quadratic formula, x = [80 ± sqrt(6400 - 1600)]/2 = [80 ± sqrt(4800)]/2. Since sqrt(4800) = 40sqrt(3), this becomes x = [80 ± 40sqrt(3)]/2 = 40 ± 20sqrt(3).
Numerically, 20sqrt(3) ≈ 34.64, so the break-even prices are about $5.36 and $74.64. Both are within the model's domain. Because the parabola opens downward and its vertex is above the x-axis, P(x) > 0 between the two roots and P(x) < 0 outside them, so the council makes a profit for prices between about $5.36 and $74.64.
3. Determine the price interval that gives a profit of at least $3000. Then test the proposal to charge $60 and state whether it meets the principal's target.
Read the worked answer
Require P(x) >= 3000:
-3x^2 + 240x - 1200 >= 3000 -3x^2 + 240x - 4200 >= 0.
Divide by -3 and reverse the inequality:
x^2 - 80x + 1400 <= 0.
The roots are x = [80 ± sqrt(6400 - 5600)]/2 = [80 ± sqrt(800)]/2 = [80 ± 20sqrt(2)]/2 = 40 ± 10sqrt(2). Numerically, 10sqrt(2) ≈ 14.14, so the interval is approximately 25.86 <= x <= 54.14. The council must charge between about $25.86 and $54.14 to reach $3000.
Test x = 60: P(60) = -3(60^2) + 240(60) - 1200 = -10800 + 14400 - 1200 = 2400. This is below $3000, so the $60 proposal does not meet the principal's target.