Ticket Sales for the Spring Show
The drama club is selling tickets for the spring show in a 300-seat auditorium. Adult tickets cost $12 and student tickets cost $7. The club needs at least $2,400 in ticket revenue to cover production costs. The sponsor also promised that student tickets will be at least twice the number of adult tickets. One volunteer proposes selling 80 adult and 200 student tickets. Another proposes 100 adult and 190 student tickets. Both volunteers claim their plan meets every condition. The club president wants to know whether either proposal actually works, and if not, what a valid ticket mix could be. The auditorium can be filled, but the club is not required to sell every seat. A square section of the stage has area 150 square feet, and the director estimates its side length as 12.2 feet. Before the meeting, the president needs a clear recommendation.
1. Test each proposal against the seating, revenue, and student-ticket conditions. Does either proposal work?
Read the worked answer
Check each proposal against three conditions: seats ≤ 300, revenue ≥ $2,400, and students ≥ 2 × adults.
Proposal A: 80 adult, 200 student. Seats = 280, revenue = 80(12) + 200(7) = 960 + 1,400 = $2,360. This is below $2,400, so it fails even though 200 ≥ 160.
Proposal B: 100 adult, 190 student. Seats = 290, revenue = 1,200 + 1,330 = $2,530, which meets the money condition. But 190 < 2(100) = 200, so it fails the student-ticket promise.
Neither proposal satisfies all conditions.
2. Find a ticket mix that satisfies all conditions, and explain how you know others exist.
Read the worked answer
Let a be adult tickets and s be student tickets. The conditions are a + s ≤ 300, 12a + 7s ≥ 2,400, and s ≥ 2a.
If the club fills all 300 seats, s = 300 - a. Revenue becomes 12a + 7(300 - a) = 2,100 + 5a. Requiring at least $2,400 gives 2,100 + 5a ≥ 2,400, so a ≥ 60. The ratio condition gives 300 - a ≥ 2a, so a ≤ 100.
Therefore any whole-number split with 60 ≤ a ≤ 100 and s = 300 - a works. Example: 80 adult and 220 student tickets gives revenue 960 + 1,540 = $2,500 and 220 ≥ 160. Many answers are valid.
3. Check the director's estimate that the square stage section with area 150 square feet has side length 12.2 feet. Is the estimate reasonable, and is it safe to use?
Read the worked answer
The square stage section has area 150 square feet, so its side length is √150. Since 12^2 = 144 and 13^2 = 169, the side is between 12 and 13 feet. More precisely, 12.2^2 = 148.84 and 12.3^2 = 151.29, so √150 ≈ 12.25 feet.
Rounded to the nearest tenth, 12.2 feet is reasonable, but it is a slight underestimate of about 0.05 foot, or about 0.6 inch. If the director needs the section to fit inside a 12.2-foot space, the estimate is not safe; if the space is at least 12.3 feet, it fits.