Isoquants and isocosts
At \(w=r=5\), the lowest isocost reaching this isoquant is \(C=40\), at \(L=K=4\).
You have two hours available on Saturday.
| Choice | Pay | Travel cost |
|---|---|---|
| Give a tutoring session | £40 | £4 |
| Work a café shift | £24 | £0 |
Assume these are your only alternatives and all other costs are zero.
How much do you gain by choosing tutoring?
Tutoring leaves you with £36 after travel: accounting profit in this simplified example.
But taking that session means giving up the £24 café shift.
\[ \text{Economic profit from tutoring} = 40 - 4 - 24 = 12\text{ pounds} \]
Economic cost includes the value of the next best alternative.
Fixed cost: independent of current output, over the period considered.
A pizza shop pays the same monthly rent whether it makes 100 or 200 pizzas today.
Sunk cost: already incurred and unrecoverable.
A £40 concert ticket you cannot refund or resell is a sunk expenditure.
A fixed input can still have an opportunity cost: an oven might be sold or rented out.
The firm uses labour \(L\) and capital \(K\).
\[ C = wL + rK \]
For a given output, choose the least costly feasible input combination.
With output price \(P\) and technology \(q=f(L,K)\), economic profit is \(\pi=Pf(L,K)-wL-rK\), holding prices and technology fixed.
A pizza kitchen is considering a small change in its inputs.
| Input | Marginal product | Input price | Output per £ |
|---|---|---|---|
| Labour-hour | 12 pizzas | £12 | 1 pizza |
| Machine-hour | 30 pizzas | £20 | 1.5 pizzas |
Here, machine time gives more output per pound.
The firm can reduce cost by using more machine time and less labour, while keeping output fixed.
At a smooth interior optimum,
\[ \frac{MP_L}{w}=\frac{MP_K}{r} \qquad\Longleftrightarrow\qquad \frac{MP_L}{MP_K}=\frac{w}{r}. \]
The last pound spent on either input yields the same marginal output.
At \(w=r=5\), the lowest isocost reaching this isoquant is \(C=40\), at \(L=K=4\).
Short run: at least one input is fixed.
A pizza shop can change staff hours today, but cannot immediately add another oven.
Long run: all inputs can adjust.
The shop can choose a different kitchen, more ovens and a different number of staff.
The distinction depends on what can change, rather than a fixed number of months.
\[ TC(q)=FC+VC(q) \]
For a pizza shop over a given day:
If \(VC(0)=0\), then \(TC(0)=FC\).
VC starts at zero. TC starts at FC. Their vertical distance is always FC.
Average costs measure cost per unit of output.
\[ AFC(q)=\frac{FC}{q},\qquad AVC(q)=\frac{VC(q)}{q} \]
\[ ATC(q)=\frac{TC(q)}{q}=AFC(q)+AVC(q) \]
Throughout the lecture, \(AC\) and \(ATC\) both mean average total cost.
Average costs are undefined at \(q=0\).
Marginal cost measures the additional cost of producing more output.
For a table with one-unit steps,
\[ MC\text{ of unit }q=TC(q)-TC(q-1). \]
For a smooth cost function,
\[ MC(q)=\frac{dTC(q)}{dq}=\frac{dVC(q)}{dq}. \]
Fixed cost does not enter MC because it does not change with output.
Your average mark across four tests is 60.
| Mark on the fifth test | New average |
|---|---|
| 50 | 58 |
| 70 | 62 |
The next observation pulls the average towards itself.
Can marginal cost be rising while average cost is falling?
MC rises throughout this example. AC falls until MC reaches AC, then rises.
\[ \frac{dAC}{dq} =\frac{q\,MC-TC}{q^2} =\frac{MC-AC}{q}. \]
The same reasoning gives \(MC=AVC\) at an interior minimum of AVC.
Here AVC is U-shaped. MC meets AVC and ATC at their respective minima.
\(ATC-AVC=AFC\) falls as output rises. The gap \(ATC-AFC=AVC\) follows AVC.
| q | FC | VC | TC | MC | AFC | AVC | ATC |
|---|---|---|---|---|---|---|---|
| 0 | 50 | 0 | 50 | — | — | — | — |
| 1 | 50 | 40 | 90 | 40 | 50.00 | 40.00 | 90.00 |
| 2 | 50 | 70 | 120 | 30 | 25.00 | 35.00 | 60.00 |
| 3 | 50 | 90 | 140 | 20 | 16.67 | 30.00 | 46.67 |
| 4 | 50 | 100 | 150 | 10 | 12.50 | 25.00 | 37.50 |
| 5 | 50 | 120 | 170 | 20 | 10.00 | 24.00 | 34.00 |
The fifth unit costs more than the fourth, but less than the previous average cost.
ATC therefore keeps falling. The table has not yet reached its turning point.
Suppose the sixth unit adds £30 to total cost, and the seventh adds £40.
| q | FC | VC | TC | MC | AVC | ATC | |
|---|---|---|---|---|---|---|---|
| 5 | 50 | 120 | 170 | 20 | 24.00 | 34.00 | |
| 6 | 50 | ? | ? | 30 | ? | ? | |
| 7 | 50 | ? | ? | 40 | ? | ? |
| q | FC | VC | TC | MC | AVC | ATC | |
|---|---|---|---|---|---|---|---|
| 5 | 50 | 120 | 170 | 20 | 24.00 | 34.00 | |
| 6 | 50 | 150 | 200 | 30 | 25.00 | 33.33 | |
| 7 | 50 | 190 | 240 | 40 | 27.14 | 34.29 |
For the sixth unit, MC = 30 is above the previous AVC of 24, but below the previous ATC of 34.
AVC rises and ATC falls. Falling AFC more than offsets rising AVC.
Q = 6 has the lowest ATC among the quantities shown.
Each extra worker-hour costs £12. The oven is fixed.
| Worker-hours | Total pizzas | Extra pizzas from the last worker-hour |
|---|---|---|
| 0 | 0 | — |
| 1 | 6 | 6 |
| 2 | 18 | 12 |
| 3 | 36 | 18 |
| 4 | 48 | 12 |
| 5 | 54 | 6 |
Calculate the labour cost per extra pizza for each step. Why does it fall and then rise?
| Last worker-hour | Extra pizzas | Extra labour cost | Labour cost per extra pizza |
|---|---|---|---|
| 1st | 6 | £12 | £2.00 |
| 2nd | 12 | £12 | £1.00 |
| 3rd | 18 | £12 | £0.67 |
| 4th | 12 | £12 | £1.00 |
| 5th | 6 | £12 | £2.00 |
Initially, workers can specialise. Eventually they must share the same oven and workspace.
Each hour costs the same. Lower extra output per hour means higher cost per extra pizza.
If labour is the only variable input, at a constant wage,
\[ VC(q)=wL(q). \]
One extra unit needs approximately \(1/MP_L\) labour-hours, so
\[ MC(q)=w\frac{dL}{dq}=\frac{w}{MP_L}. \]
With capital fixed, diminishing \(MP_L\) raises MC.
For the pizza shop, consider each change separately.
Which of TC, AFC, AVC, ATC and MC change?
| Change | TC | AFC | AVC | ATC | MC |
|---|---|---|---|---|---|
| Rent +£100 | +100 | +100/q | Same | +100/q | Same |
| Ingredients +50p per pizza | +0.50q | Same | +0.50 | +0.50 | +0.50 |
| Higher demand alone | Same curve | Same | Same | Same | Same |
Higher demand may change the output the firm chooses. This moves it along its existing cost curves.
Changes in input prices or technology change the cost curves themselves.
An independent studio spends £100,000 developing a game.
Each copy costs another £2 to distribute and support. Assume these are its only costs.
\[ TC(q)=100{,}000+2q. \]
| Copies sold | Total cost | Average cost | Marginal cost |
|---|---|---|---|
| 10,000 | £120,000 | £12 | £2 |
| 50,000 | £200,000 | £4 | £2 |
\[ AC(q)=\frac{100{,}000}{q}+2,\qquad MC(q)=2. \]
AC falls because development cost is spread across more copies. MC stays constant.
Costs alone do not tell us the best selling price. That also depends on demand and competition.
Every row produces 40 burgers per hour. Labour and grill services each cost £5 per unit.
| Labour L | Grill services K | Cost = 5L + 5K |
|---|---|---|
| 1 | 16 | £85 |
| 2 | 8 | £50 |
| 4 | 4 | £40 |
| 8 | 2 | £50 |
| 10 | 1.6 | £58 |
Which mix is cheapest? Why can using fewer workers cost more?
The production function tells us how many burgers the inputs can make:
\[ q=10\sqrt{LK}. \]
To make 40 burgers per hour, divide by 10 and square:
\[ 40=10\sqrt{LK}\quad\Rightarrow\quad 4=\sqrt{LK} \quad\Rightarrow\quad LK=16. \]
Both \((L,K)=(2,8)\) and \((4,4)\) work because both products equal 16.
Each input costs £5 per unit, so the first mix costs \(5(2)+5(8)=50\) pounds per hour and the second costs \(5(4)+5(4)=40\).
\[ q=10\sqrt{4L}=20\sqrt L. \]
\[ \frac{q}{20}=\sqrt L \quad\Rightarrow\quad L=\left(\frac{q}{20}\right)^2=\frac{q^2}{400}. \]
\[ STC(q)=5(4)+5\left(\frac{q^2}{400}\right)=20+\frac{q^2}{80}. \]
At \(q=20\), one unit of labour is enough: \(L=(20/20)^2=1\). Total cost is \(5(4)+5(1)=25\) pounds per hour.
Divide total cost by \(q\) to get average cost. Differentiate total cost to get marginal cost:
\[ \begin{aligned} SAC(q)&=\frac{20+q^2/80}{q}=\frac{20}{q}+\frac{q}{80},\\[0.25em] SMC(q)&=\frac{d}{dq}\left(20+\frac{q^2}{80}\right)=\frac{2q}{80}=\frac{q}{40}. \end{aligned} \]
| Burgers per hour | STC (£ per hour) | SAC (£ per burger) | SMC (£ per burger) |
|---|---|---|---|
| 20 | £25 | £1.25 | £0.50 |
| 40 | £40 | £1.00 | £1.00 |
| 80 | £100 | £1.25 | £2.00 |
\[ \frac{MP_L}{MP_K}=\frac{K}{L}=\frac{5}{5}=1 \quad\Rightarrow\quad L=K. \]
\[ q=10\sqrt{L^2}=10L \quad\Rightarrow\quad L=K=\frac{q}{10}. \]
Dividing by \(q\) gives \(LRAC=1\). Differentiating gives \(LRMC=1\).
Four grills are exactly right at q = 40. At other outputs, changing equipment can reduce cost.
| q | Best long-run (L, K) | STC with K = 4 | LRTC |
|---|---|---|---|
| 20 | (2, 2) | £25 | £20 |
| 40 | (4, 4) | £40 | £40 |
| 80 | (8, 8) | £100 | £80 |
\[ STC(q)-LRTC(q)=\frac{(q-40)^2}{80}\geq0. \]
Short-run choice can be optimal given fixed capital. Allowing capital to change weakly lowers minimum cost.
The expansion path traces cost-minimising input combinations as output changes, holding input prices fixed.
Diminishing marginal product
Increase labour while holding equipment fixed.
Returns to scale
Increase all inputs in the same proportion.
In Hamburger Heaven, \(q=10\sqrt{LK}\):
Both statements hold for the same technology.
Suppose all inputs double and their prices stay fixed. Spending on that input bundle doubles.
| Output from the scaled bundle | Average cost relative to before |
|---|---|
| Output doubles | Unchanged |
| Output more than doubles | Lower |
| Output less than doubles | Higher |
With homogeneous technologies, fixed input prices and cost minimisation, these give the familiar link between returns to scale and long-run average cost.
Increasing returns correspond to falling LRAC. Decreasing returns correspond to rising LRAC, under the stated assumptions.
Economies of scale: long-run average cost falls as output increases.
Possible mechanisms include specialisation and indivisible equipment.
Diseconomies of scale: long-run average cost rises as output increases.
Coordination and management can become more difficult as an organisation grows.
Minimum efficient scale: the smallest output attaining minimum LRAC, when that minimum exists.
For each output, LRAC selects the least costly plant from all available sizes. Only three SRAC curves are shown.
At the same output, long-run minimum cost cannot exceed short-run minimum cost.
The short-run plan remains available when all inputs can adjust.
With a smooth range of plant sizes, LRAC is tangent to the selected SRAC.
That tangency need not be the minimum of the selected SRAC.
With only a few plant sizes available, the lower envelope can have kinks.
A technology improvement expands what the firm can produce with its inputs. Minimum cost at a given output weakly falls.
Economies of scope: producing two products jointly is cheaper than producing the same quantities separately.
A café can share its premises and till between coffee and sandwiches.
\[ C(q_1,q_2)<C(q_1,0)+C(0,q_2). \]
Scale concerns output size. Scope concerns joint production across products.
Suppose
\[ TC(q)=100+q+0.01q^2. \]
\[ AFC=\frac{100}{q},\quad AVC=1+0.01q, \quad ATC=\frac{100}{q}+1+0.01q. \]
\[ MC=1+0.02q,\qquad \frac{dATC}{dq}=-\frac{100}{q^2}+0.01=0. \]
Minimum ATC occurs at q = 100, where ATC = MC = £3.
Choosing the profit-maximising output also requires information about revenue.
Give a reason for each answer.
The recurring question is what can change, and what are we holding fixed?