Question 3 OR (a)
Q = AL^(3/8)K^(3/8): returns to scale, returns to each input and the adding-up result
9 marks
- Time to write
- About 18 minutes
Question
Question 3 OR (a) · 9 marksA production function is given by Q = AL^{3/8}K^{3/8}, where L is labour and K is capital.
What is the nature of returns to scale?
What is the nature of returns to inputs?
When each factor is paid a price equal to its marginal product, what is the total reward of labour and capital? Show that total product is not exhausted.
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.
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- What is the…?When…?Show…4 questions in all: answer every one.
- Show thatThe result is given: the marks are for the steps that reach it.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Returns to scale 3 marks 5 min
- 2 Returns to inputs 3 marks 5 min
- 3 Total reward and non-exhaustion 3 marks 5 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
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Solution
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The production function is of the Cobb-Douglas form, as the DU guideline uses for this topic.
Q(tL, tK) = A(tL)^(3/8)(tK)^(3/8) = t^(3/8 + 3/8) A L^(3/8)K^(3/8) = t^(3/4) Q.
The function is homogeneous of degree 3/4 < 1, so it shows decreasing returns to scale: doubling both inputs raises output by the factor 2^(3/4) ≈ 1.68, less than double.
(ii) Returns to each input (returns to a factor)
MP_L = ∂Q/∂L = (3/8)A L^(−5/8)K^(3/8) = (3/8)Q/L > 0 ∂²Q/∂L² = (3/8)(−5/8)A L^(−13/8)K^(3/8) = −(15/64)Q/L² < 0
MP_K = ∂Q/∂K = (3/8)Q/K > 0, ∂²Q/∂K² = −(15/64)Q/K² < 0.
Each marginal product is positive but falls as more of that input is used with the other held fixed: diminishing returns to each input (the exponent 3/8 is below 1).
(iii) Total reward when each factor is paid its marginal product
L·MP_L + K·MP_K = L(3/8)(Q/L) + K(3/8)(Q/K) = (3/8)Q + (3/8)Q = (3/4)Q.
Labour receives 3/8 of the product and capital 3/8; together they receive (3/4)Q, which is less than Q. So the total product is not exhausted: Q − (3/4)Q = (1/4)Q remains after the factors are paid their marginal products (Euler's theorem for a function homogeneous of degree 3/4).
Numerical illustration (A = 10, L = 256, K = 1): Q = 10 × 256^(3/8) = 10 × 8 = 80; MP_L = (3/8)(80/256) = 0.1171875, L·MP_L = 30; MP_K = (3/8)(80) = 30, K·MP_K = 30; total reward 60 = (3/4) × 80, leaving 20 unpaid.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Returns to scale | 3 |
| 2 | Returns to inputs | 3 |
| 3 | Total reward and non-exhaustion | 3 |
| Total | 9 |
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