Question 3 (b)

Demand function from its price elasticity

9 marks

Time to write
About 18 minutes
Topic
Integration: elasticities and surplus

Question

Question 3 (b) · 9 marks

The price elasticity of demand for a commodity is Eₚ = P/[(P - 2)(P - 3)], where P is the price per unit. Obtain the demand function that corresponds to a demand of 6 units when the price is ₹4. (9)

The same question was set as 832-Q3b; this answer serves both.

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

18 minfor 9 marks

Write it in this order

  1. Read the data; note exactly what is asked 1 min
  2. 1 Differential equation from the elasticity 2 marks 4 min
  3. 2 Partial fractions and integration 4 marks 7 min
  4. 3 Constant from x = 6 at p = 4 2 marks 3 min
  5. 4 Demand function with check 1 mark 2 min
  6. 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 p458
Quantity x = 3(p − 2)/(p − 3)64.53.6
Elasticity p/[(p − 2)(p − 3)]20.830.27

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

#ComponentMarks
1Differential equation from the elasticity2
2Partial fractions and integration4
3Constant from x = 6 at p = 42
4Demand function with check1
Total9
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