Question 2 (a)
OTT subscription price that maximises revenue
9 marks
- Time to write
- About 18 minutes
Question
Question 2 (a) · 9 marksAn OTT service operator has 1,500 subscribers, each of whom he charges ₹400 per month. To increase his subscriber base, he proposes to lower the rate of monthly subscriptions. He assesses that for every decrease of ₹2 in the monthly subscription, ten more subscribers would seek the OTT service. You are required to:
Compute the fall in the service charge and the new number of subscribers that will let him maximize his total revenue.
Find the corresponding new revenue.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- Compute…Find…Two asks in one question: answer both.
- ComputeFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Revenue function 3 marks 5 min
- 2 First- and second-order conditions 3 marks 5 min
- 3 Fall in charge and new subscribers 2 marks 4 min
- 4 New revenue 1 mark 2 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Revenue function. Let x be the number of ₹2 reductions in the monthly charge. Then
Monthly charge per subscriber = 400 − 2x (₹); number of subscribers = 1500 + 10x;
Total revenue R(x) = (400 − 2x)(1500 + 10x) = 6,00,000 + 4,000x − 3,000x − 20x² = 6,00,000 + 1,000x − 20x².
dR/dx = 1,000 − 40x; setting dR/dx = 0 gives x = 25.
d²R/dx² = −40 < 0, so x = 25 gives a maximum (second derivative test). The first and second derivatives follow the sum and power rules of differentiation.
(i) Fall in the charge and new subscribers
Fall in the monthly charge = 2 × 25 = ₹50, so the new charge is 400 − 50 = ₹350 per month.
New number of subscribers = 1500 + 10 × 25 = 1,750.
R = 350 × 1,750 = ₹6,12,500 per month, against the present 400 × 1,500 = ₹6,00,000, an increase of ₹12,500.
| Present | Revenue-maximising | |
|---|---|---|
| Monthly charge (₹) | 400 | 350 |
| Subscribers | 1,500 | 1,750 |
| Monthly revenue (₹) | 6,00,000 | 6,12,500 |
Check: at x = 24 (₹352, 1,740 subscribers) R = ₹6,12,480 and at x = 26 (₹348, 1,760) R = ₹6,12,480, both below ₹6,12,500.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Revenue function | 3 |
| 2 | First- and second-order conditions | 3 |
| 3 | Fall in charge and new subscribers | 2 |
| 4 | New revenue | 1 |
| Total | 9 |
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