Question 2 OR (a)
Price ranges of elastic and inelastic demand, and the revenue-maximising price
9 marks
- Time to write
- About 18 minutes
Question
Question 2 OR (a) · 9 marksThe demand function for a certain commodity is 4 q = 600 - 3 p (0 ≤ p ≤ 200).
Determine the price ranges over which demand is elastic, inelastic or of unit elasticity.
Use the result of part (i) to find the intervals over which the revenue function increases and decreases, and the price that maximizes 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
- DetermineFormula 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 Elasticity function 2 marks 4 min
- 2 Elastic, inelastic and unit-elastic ranges 3 marks 5 min
- 3 Revenue increasing and decreasing intervals 3 marks 5 min
- 4 Revenue-maximising price 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.
Demand: 4q = 600 − 3p, i.e. q = 150 − 0.75p, for 0 ≤ p ≤ 200. dq/dp = −0.75.
(i) Price elasticity of demand
η = −(dq/dp)(p/q) = 0.75p/(150 − 0.75p) = 3p/(600 − 3p) = p/(200 − p).
- Unit elasticity: η = 1 ⟹ p = 200 − p ⟹ p = 100.
- Elastic (η > 1): p > 200 − p ⟹ 100 < p ≤ 200 (η grows without bound as p → 200, where q = 0).
- Inelastic (η < 1): 0 ≤ p < 100 (η = 0 at p = 0).
| Price p | 0 | 50 | 100 | 150 |
|---|---|---|---|---|
| Quantity q | 150 | 112.5 | 75 | 37.5 |
| Elasticity η | 0 | 0.33 | 1 | 3 |
| Revenue pq | 0 | 5,625 | 7,500 | 5,625 |
Revenue R = pq = 150p − 0.75p², so dR/dp = 150 − 1.5p = q(1 − η) (since dR/dp = q + p·dq/dp = q[1 − η]).
- Where demand is inelastic (0 ≤ p < 100), η < 1, so dR/dp > 0: revenue increases as price rises.
- Where demand is elastic (100 < p ≤ 200), η > 1, so dR/dp < 0: revenue decreases as price rises.
- At p = 100 (unit elasticity) dR/dp = 0 and d²R/dp² = −1.5 < 0, so revenue is a maximum: q = 75 and R = 100 × 75 = ₹7,500.
The revenue-maximising price is ₹100.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Elasticity function | 2 |
| 2 | Elastic, inelastic and unit-elastic ranges | 3 |
| 3 | Revenue increasing and decreasing intervals | 3 |
| 4 | Revenue-maximising price | 1 |
| Total | 9 |
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