Difficulty: Intermediate | Prerequisites: Production functions, marginal products, MRTS (see Part I study notes).
Cost minimisation asks: given a target output level, what combination of inputs costs the least? For Cobb-Douglas functions the answer is a tangency condition (MRTS = input price ratio). For Leontief functions, inputs must be used in fixed proportions, so you solve the min condition and work out quantities directly. This is the bridge between production theory and the cost curves you will need for market analysis.
Isocost line
A line showing all combinations of two inputs that cost the same total amount. The equation is C = wL + rK, where w is the price of labour (or input 1) and r is the price of capital (or input 2). Rearranging: K = C/r - (w/r) · L.
Think of it as: the budget line for a firm, analogous to the consumer's budget constraint.
Cost minimisation
Choosing the input bundle that produces a given output Q at the lowest total cost. The firm picks the point where the isoquant is tangent to the lowest possible isocost line.
In simple terms: you want to hit your production target while spending as little as possible.
Tangency condition
At the cost-minimising point, MRTS = w/r (the ratio of input prices). Equivalently, MP_L/w = MP_K/r, meaning the last pound spent on each input yields the same marginal output.
Think of it as: if one input gives you more bang per pound than the other, you have not yet minimised cost.
Leontief (fixed-proportions) optimisation
When the production function is Q = min{aX₁, bX₂}, cost minimisation requires aX₁ = bX₂ (use inputs in exactly the required ratio, no waste). Any excess of one input is wasted money.
In simple terms: match the binding constraints. If R = min{f, w}, you need f = w. Any extra fertiliser or water beyond that ratio is money down the drain.
Willingness to pay for technology
The maximum a firm would pay for a new technology equals the cost saving it produces: old cost minus new cost at the same output level.
Think of it as: the technology is worth whatever it saves you.
Normal conditions: R = min{f, w} with r_f = $3, r_w = $2.
For 1 unit of rice, the binding constraint is f = w (use both in equal proportion). So f* = 1, w* = 1. Total cost = 3(1) + 2(1) = $5.
Drought conditions: R = min{f, 0.2w}.
Now 0.2w must equal f for no waste. For R = 1: f = 1 and 0.2w = 1, so w = 5. Total cost = 3(1) + 2(5) = $13.
Drought makes rice dramatically more expensive because you need five times as much water.
New seed technology: R = min{3f, 0.2w}.
Now 3f = 0.2w for no waste. For R = 1: 3f = 1 gives f = 1/3, and 0.2w = 1 gives w = 5. Total cost = 3(1/3) + 2(5) = 1 + 10 = $11.
The saving from the new technology is $13 - $11 = $2. That is the most the farmer would pay for it.
Setup: Q = 2L^(1/4) · K^(1/4), w = $32, r = $2, total cost C = $64.
Step 1: Find marginal products.
MP_L = 2 · (1/4) · L^(-3/4) · K^(1/4) = (1/2) · L^(-3/4) · K^(1/4)
MP_K = 2 · (1/4) · L^(1/4) · K^(-3/4) = (1/2) · L^(1/4) · K^(-3/4)
Step 2: Apply the tangency condition.
MRTS = MP_L / MP_K = K/L. Set MRTS = w/r: K/L = 32/2 = 16, so K = 16L.
Step 3: Substitute into the cost constraint.
32L + 2(16L) = 64, so 32L + 32L = 64, giving L = 1 and K = 16.
Step 4: Find output.
Q = 2(1)^(1/4) · (16)^(1/4) = 2 · 1 · 2 = 4 units.
Same technology, same price ratio, so K = 16L still holds.
Plug into the production function: 2L^(1/4) · (16L)^(1/4) = 2, which gives 2 · 2 · L^(1/2) = 2, so L^(1/2) = 1/2, L = 1/4, K = 4.
New cost = 32(1/4) + 2(4) = 8 + 8 = $16.
Halving output from Q = 4 to Q = 2 reduced cost from $64 to $16, a 75% cut. Output fell by half, but cost fell by three-quarters. This tells you the function has increasing returns to scale (confirmed: exponent sum = 1/4 + 1/4 = 1/2 < 1, which is actually decreasing returns to scale).
Wait, let us be precise. The exponents sum to 1/2, which means decreasing returns to scale. Scaling inputs by t multiplies output by t^(1/2), less than t. This means larger scale is less efficient per unit of input, but the cost comparison works differently because of the specific numbers.
At Q = 4: revenue = 4 × $20 = $80, cost = $64, profit = $16.
At Q = 2: revenue = 2 × $20 = $40, cost = $16, profit = $24.
Profit rises from $16 to $24 when the CEO cuts output. The CEO's decision is correct: despite producing fewer goods, the firm is more profitable because costs fall proportionally more than revenue under decreasing returns to scale. With DRS, the firm was over-extended at Q = 4.
Isocost line
C = wL + rK, or equivalently K = C/r - (w/r) · L.
Tangency condition (Cobb-Douglas)
MRTS = w/r, which for Q = A · L^α · K^β gives (α/β) · (K/L) = w/r, so K = (βw / αr) · L.
Leontief cost minimisation
For R = min{af, bw}: set af = bw, solve for one input in terms of the target output, then find the other. Cost = r_f · f* + r_w · w*.
Willingness to pay for technology
WTP = Cost_old - Cost_new (both at the same output level).
Cost minimisation is the logic behind every sourcing decision a firm makes: whether to hire more workers or invest in machinery, whether to use a cheaper but less efficient supplier. In agriculture, the Leontief framework captures real irrigation constraints: a drought that changes the water-to-output ratio immediately changes the optimal spending mix and total cost, exactly as modelled in the rice-farming problem.
Students set MRTS = w/r but forget to also use the production function or cost constraint to pin down the actual quantities. The tangency condition alone gives you a ratio (K = 16L, say), not the levels. You need a second equation.
With Leontief functions, students sometimes try to set MRTS = w/r. There is no smooth trade-off between inputs in a Leontief function. The isoquants are L-shaped, so the tangency approach does not apply. Instead, set the arguments of the min function equal to each other.
A common error on the CEO problem is to assume that cutting output always cuts profit. It depends on returns to scale and the price level. With decreasing returns, a smaller firm can be more profitable.
Students confuse the cost constraint (C = wL + rK) with the production constraint (Q = f(L,K)). In cost minimisation with a fixed budget, you use both. In cost minimisation for a target output, you use the tangency condition plus the production function.
⚠️ The tangency condition (MRTS = w/r) will almost certainly appear on the exam. Be ready to derive it from scratch for any Cobb-Douglas function.
⚠️ Leontief cost minimisation is straightforward but easy to rush. Write out the no-waste condition (set the arguments of min equal) and solve carefully. The numbers change when the coefficients inside the min function change (as in the drought scenario).
⚠️ Willingness to pay for technology is a favourite follow-up question. It is always the cost difference at the same output.
⚠️ The returns-to-scale question at the end connects production theory to profit. Know how to compute profit (revenue minus cost) and reason about whether scaling up or down is beneficial.
True or false: For a Leontief production function R = min{f, w}, the cost-minimising bundle always has f = w. ___
Fill in the blank: The tangency condition for cost minimisation is MRTS = ___.
True or false: If a firm has decreasing returns to scale, cutting output can increase profit. ___
Fill in the blank: The willingness to pay for a cost-saving technology equals ___ minus ___.
True or false: For Q = 2L^(1/4) · K^(1/4) with w = 32 and r = 2, the optimal ratio is K = 16L. ___
Answers: 1. True (the no-waste condition requires both arguments equal). 2. w/r (the input price ratio). 3. True. 4. Old cost minus new cost (at the same output). 5. True (from K/L = w/r = 16).
Q: A farmer's production function is R = min{f, 0.2w} with r_f = $3 and r_w = $2. What is the cost-minimising bundle for 1 unit of R?
A: Set f = 0.2w. For R = 1: f = 1, so 0.2w = 1, w = 5. Cost = 3(1) + 2(5) = $13.
Q: For Q = 2L^(1/4) · K^(1/4) with w = $32 and r = $2, find the cost-minimising L, K, and Q when total cost is $64.
A: Tangency: K/L = 32/2 = 16, so K = 16L. Cost constraint: 32L + 2(16L) = 64, giving L = 1, K = 16. Q = 2(1)(16)^(1/4) = 2 × 2 = 4.
Q: Same firm, but now the target is Q = 2. Find the new L, K, and cost.
A: K = 16L still holds. From 2L^(1/4)(16L)^(1/4) = 2: 4L^(1/2) = 2, L = 1/4, K = 4. Cost = 32(1/4) + 2(4) = $16.
Q: At Q = 4 (cost $64, revenue $80) and Q = 2 (cost $16, revenue $40), which output level gives higher profit?
A: Profit at Q = 4 is $80 - $64 = $16. Profit at Q = 2 is $40 - $16 = $24. Q = 2 is more profitable. The CEO's cost-cutting decision is correct.
Q: What is the maximum the rice farmer would pay for the new seed technology (R = min{3f, 0.2w}) during a drought?
A: Drought cost without new seed: $13. With new seed: $11. Maximum willingness to pay = $13 - $11 = $2.
Cost minimisation is the supply-side twin of utility maximisation from consumer theory: same tangency logic, different labels (MRTS instead of MRS, isocost instead of budget line). The optimal input bundles derived here feed directly into cost functions and eventually into supply curves. Returns to scale determines the shape of the long-run average cost curve: decreasing returns produce an upward-sloping LRAC, which matters for predicting firm behaviour in competitive markets.
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