Question 3 (b)
Demand function from its price elasticity
9 marks
- Time to write
- About 18 minutes
Question
Question 3 (b) · 9 marksThe elasticity of demand for a commodity with respect to price is found to be p/((p - 2)(p - 3)), where p is the price. Obtain the demand function, given that the quantity demanded is 6 units at p = 4.
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
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Differential equation from the elasticity 2 marks 4 min
- 2 Partial fractions and integration 4 marks 7 min
- 3 Constant from x = 6 at p = 4 2 marks 3 min
- 4 Demand function with check 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.
Setting up. Price is written p, quantity x and the price elasticity η (Eₚ). The price elasticity of demand is η = −(p/x)(dx/dp), the positive number for a downward-sloping demand curve. Given η = p/[(p − 2)(p − 3)]:
−(p/x)(dx/dp) = p/[(p − 2)(p − 3)] ⟹ dx/x = −dp/[(p − 2)(p − 3)].
Integrating (deriving the demand function from the elasticity of demand, a Unit 3 integration topic). By partial fractions, 1/[(p − 2)(p − 3)] = 1/(p − 3) − 1/(p − 2). Hence, using ∫dx/x = log x,
log x = −log(p − 3) + log(p − 2) + log k ⟹ x = k(p − 2)/(p − 3) (for p > 3).
Constant from the given point. x = 6 when p = 4: 6 = k(2)/(1) ⟹ k = 3.
Demand function: x = 3(p − 2)/(p − 3), p > 3.
Check. dx/dp = 3[(p − 3) − (p − 2)]/(p − 3)² = −3/(p − 3)² < 0, so demand falls as price rises, and
−(p/x)(dx/dp) = [p(p − 3)/(3(p − 2))] × [3/(p − 3)²] = p/[(p − 2)(p − 3)], the given elasticity. At p = 4: x = 3 × 2/1 = 6.
| Price p | 4 | 5 | 8 |
|---|---|---|---|
| Quantity x = 3(p − 2)/(p − 3) | 6 | 4.5 | 3.6 |
| Elasticity p/[(p − 2)(p − 3)] | 2 | 0.83 | 0.27 |
The other reading. If the elasticity were defined without the minus sign, (p/x)(dx/dp) = p/[(p − 2)(p − 3)], the same steps give x = 12(p − 3)/(p − 2), which rises with price; it is not a demand law, so the signed (negative-slope) reading above is taken.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Differential equation from the elasticity | 2 |
| 2 | Partial fractions and integration | 4 |
| 3 | Constant from x = 6 at p = 4 | 2 |
| 4 | Demand function with check | 1 |
| Total | 9 |
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