Vista Viewing Centre: Snack Prices and the Best Ticket Price
The Vista Viewing Centre in Enugu has 300 seats and charges ₦500 per match ticket. On a typical match day 240 seats are sold. The manager's survey of regular customers suggests that for every ₦25 added to the ticket price, 10 fewer seats would be sold, and this pattern is said to hold up to a ₦250 increase. Inside the hall a kiosk sells meat pies and bottled water. On the first Saturday the kiosk sold 120 pies and 80 bottles and took ₦32,000; on the second Saturday it sold 90 pies and 110 bottles and took ₦33,500. Prices did not change between the two Saturdays. The centre must raise gate income to at least ₦115,000 per match to service a new loan, and the manager wants both the snack prices verified and the ticket price set so that gate income is as high as possible without dropping below that minimum.
1. Form and solve simultaneous equations to find the price of a meat pie and the price of a bottle of water.
Read the worked answer
Let p = price of a pie and w = price of a bottle, in naira.
Day 1: 120p + 80w = 32,000 → divide by 40 → 3p + 2w = 800 Day 2: 90p + 110w = 33,500 → divide by 10 → 9p + 11w = 3,350
From the first equation, 3p = 800 − 2w, so 9p = 2,400 − 6w. Substituting into the second: 2,400 − 6w + 11w = 3,350 → 5w = 950 → w = ₦190. Then 3p = 800 − 2(190) = 420 → p = ₦140.
Check in the original data: Day 1: 120(140) + 80(190) = 16,800 + 15,200 = ₦32,000 ✓. Day 2: 90(140) + 110(190) = 12,600 + 20,900 = ₦33,500 ✓.
The check matters: simplifying by dividing only one equation, or eliminating the wrong variable, is the usual source of a plausible but wrong pair such as ₦150 and ₦180.
2. Let x be the number of ₦25 increases. Form the gate-income function and find the ticket price that maximises it, stating the maximum income.
Read the worked answer
Price = 500 + 25x; attendance = 240 − 10x (with 0 ≤ x ≤ 10 from the survey, and attendance never exceeding the 300 seats).
R(x) = (500 + 25x)(240 − 10x) = 120,000 − 5,000x + 6,000x − 250x² = 120,000 + 1,000x − 250x²
Completing the square: R(x) = −250(x² − 4x) + 120,000 = −250[(x − 2)² − 4] + 120,000 = 121,000 − 250(x − 2)².
Since (x − 2)² ≥ 0, R is largest when x = 2, giving price 500 + 50 = ₦550, attendance 240 − 20 = 220, and income 550 × 220 = ₦121,000.
Judgement. The gain over the present ₦120,000 is only ₦1,000 (about 0.8%), while the price rises 10% and 20 customers are lost. The maximum is real but thin, so the manager should weigh the goodwill cost; a candidate who answers 'x = 2' without comparing with the present income has not finished the question.
3. Using the model, find the whole-number values of x for which gate income is at least ₦115,000, and state the highest ticket price that satisfies the loan condition.
Read the worked answer
Solve 120,000 + 1,000x − 250x² ≥ 115,000.
Rearrange: −250x² + 1,000x + 5,000 ≥ 0. Divide by −250 and reverse the inequality: x² − 4x − 20 ≤ 0
Roots: x = [4 ± √(16 + 80)] ÷ 2 = [4 ± √96] ÷ 2 = (4 ± 9.798) ÷ 2, so x ≈ 6.899 or x ≈ −2.899.
The inequality holds between the roots: −2.899 ≤ x ≤ 6.899. Since x must be a whole number of increases and x ≥ 0, the valid values are x = 0, 1, 2, 3, 4, 5, 6.
Test the boundary. x = 6: price ₦650, attendance 180, income = ₦117,000 ≥ ₦115,000 ✓. x = 7: price ₦675, attendance 170, income = ₦114,750 < ₦115,000 ✗.
So the highest permissible price is ₦650, which also exceeds the true maximum of ₦121,000 at ₦550 - a reminder that 'meets the loan condition' and 'maximises income' are different questions with different answers.