Source: Lecture 10 – Production
Difficulty: Intermediate | Companion to: Production Theory Study Notes
A condensed reference for revision and exam prep. Keep this open alongside the full study notes.
\text{Cobb-Douglas:} \quad f(K, L) = A K^{a} L^{b}\text{Leontief:} \quad f(K, L) = \min\{K/a_K,\; L/a_L\}\text{Perfect substitutes:} \quad f(K, L) = K + LMP_K = \frac{\partial f}{\partial K} \qquad MP_L = \frac{\partial f}{\partial L}For f = AK^a L^b specifically:
MP_K = a A K^{a-1} L^{b} \qquad MP_L = b A K^{a} L^{b-1}TRS = \frac{MP_K}{MP_L} = \frac{a}{b} \cdot \frac{L}{K} \quad \text{(for Cobb-Douglas)}Condition | Type | Meaning |
|---|---|---|
a + b = 1 | Constant (CRS) | Double inputs, double output |
a + b > 1 | Increasing (IRS) | Double inputs, more than double output |
a + b < 1 | Decreasing (DRS) | Double inputs, less than double output |
Consumer side | Producer side |
|---|---|
Utility function U(x, y) | Production function f(K, L) |
Indifference curve | Isoquant |
Marginal rate of substitution (MRS) | Technical rate of substitution (TRS) |
Marginal utility (MU) | Marginal product (MP) |
Budget constraint | Isocost line |
Maximise utility | Maximise output (or minimise cost) |
Diminishing marginal utility | Diminishing marginal product |
Step 1 – Marginal products.
Take the partial derivative with respect to each input:
MP_K = (1/2) K^(-1/2) L^(1/2) = (1/2)(L/K)^(1/2)
MP_L = (1/2) K^(1/2) L^(-1/2) = (1/2)(K/L)^(1/2)
Step 2 – TRS.
Divide MP_K by MP_L:
TRS = [(1/2)(L/K)^(1/2)] / [(1/2)(K/L)^(1/2)] = L/K
At K = 10, L = 10: TRS = 10/10 = 1. One-for-one substitution.
Step 3 – Returns to scale.
Exponents: 1/2 + 1/2 = 1. Constant returns to scale.
Verify: f(tK, tL) = (tK)^(1/2)(tL)^(1/2) = t^(1/2) · t^(1/2) · K^(1/2) L^(1/2) = t · f(K, L). Confirmed.
Step 1 – Marginal products.
MP_K = 3 · (2/3) · K^(-1/3) L^(1/2) = 2 K^(-1/3) L^(1/2)
MP_L = 3 · (1/2) · K^(2/3) L^(-1/2) = (3/2) K^(2/3) L^(-1/2)
Step 2 – TRS.
TRS = MP_K / MP_L = [2 K^(-1/3) L^(1/2)] / [(3/2) K^(2/3) L^(-1/2)]
Simplify: TRS = (2 / (3/2)) · (L^(1/2) · L^(1/2)) / (K^(-1/3) · K^(2/3))^(-1)
More carefully: TRS = (4/3) · (L/K)
Note the coefficient (a/b) = (2/3)/(1/2) = 4/3 appearing in front of L/K. This is why the general formula TRS = (a/b)(L/K) matters.
Step 3 – Returns to scale.
Exponents: 2/3 + 1/2 = 4/6 + 3/6 = 7/6 > 1. Increasing returns to scale.
Verify: f(tK, tL) = 3(tK)^(2/3)(tL)^(1/2) = 3 · t^(2/3) · t^(1/2) · K^(2/3) L^(1/2) = t^(7/6) · f(K, L). Since 7/6 > 1, output more than scales with inputs.
Production function | Isoquant shape | Why | TRS behaviour |
|---|---|---|---|
Cobb-Douglas: AK^a L^b | Smooth, convex curve bowing toward origin | Inputs are substitutable but imperfectly; diminishing marginal products make the curve steeper at the ends | Diminishing: TRS = (a/b)(L/K), falls as you move right along the curve |
Perfect substitutes: K + L | Straight line, slope = -1 | One unit of K always replaces exactly one unit of L, regardless of the mix | Constant: TRS = 1 everywhere |
Leontief: min{K, L} | L-shaped with a right-angle kink | Inputs must be used in fixed proportions; extra units of one input alone add nothing to output | Undefined at the kink; zero on horizontal segment, infinite on vertical segment |
If the question gives a Cobb-Douglas function and asks you to sketch the isoquant, draw a smooth downward-sloping convex curve. Label the axes K (vertical) and L (horizontal).
If the function is additive (K + L or aK + bL), draw a straight line. The slope is -a/b.
If the function uses "min", draw an L-shape. The kink sits on the ray from the origin where inputs are in the required ratio.
Remember: isoquants further from the origin represent higher output levels. They never cross, for the same reason indifference curves never cross.