Question 2 (b)
Cable subscription charge that maximises revenue
9 marks
- Time to write
- About 18 minutes
Question
Question 2 (b) · 9 marksA cable operator has a subscriber base of 1200 households, each charged ₹400 per month. To increase this base, he proposes to reduce the rate of monthly subscriptions. He assesses that for every reduction of ₹40, 300 more households would seek a cable connection. Find the decrease in the service charge that lets him maximize his total 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
- FindFormula 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 Decrease in charge with new charge and revenue 3 marks 5 min
- Check: arithmetic, units (₹, %), final answer underlined 2 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 ₹40 reductions in the monthly charge. Then
charge = 400 − 40x (₹), households = 1200 + 300x,
R(x) = (400 − 40x)(1200 + 300x) = 4,80,000 + 1,20,000x − 48,000x − 12,000x² = 4,80,000 + 72,000x − 12,000x².
dR/dx = 72,000 − 24,000x = 0 ⟹ x = 3; d²R/dx² = −24,000 < 0, so R is a maximum (derivatives by the power and sum rules).
Decrease in the charge = 3 × 40 = ₹120, so the new charge is ₹280 a month.
| Present | Revenue-maximising | |
|---|---|---|
| Monthly charge (₹) | 400 | 280 |
| Households | 1,200 | 2,100 |
| Monthly revenue (₹) | 4,80,000 | 5,88,000 |
Revenue rises from ₹4,80,000 to ₹5,88,000 a month (households 1,200 + 300 × 3 = 2,100).
Same result with the reduction r (₹) as the variable: households = 1200 + 7.5r, R = (400 − r)(1200 + 7.5r), dR/dr = 1,800 − 15r = 0 ⟹ r = ₹120.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Revenue function | 3 |
| 2 | First- and second-order conditions | 3 |
| 3 | Decrease in charge (with new charge and revenue) | 3 |
| Total | 9 |
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