CAPITAL here is not money: it is machines, buildings, roads, computers — everything that has been produced and that serves to produce. Write K for that stock for the whole country, L for the number of workers, and **k = K/L** for capital per worker. Thanks to constant returns to scale (assumption 1), output per worker depends on it alone. The algebra of that step takes one line: if doubling both factors doubles output, then dividing both by L divides output by L — that is, F(K, L)/L = F(K/L, 1). The right-hand side no longer depends on anything but k, and we write it f(k). Hence **y = f(k)**. We shall take f(k) = √k, a square root: it rises, but more and more slowly — going from 1 to 4 machines takes output from 1 to 2, but going from 4 to 9 takes it only from 2 to 3. There are the diminishing returns, in two numbers. (The general form is written k^α; α measures the share of capital in value added, about 1/3 in the data. We take α = 1/2 so that every computation comes out round, and the simulator will let you test 1/3.)

What becomes of the capital stock from one year to the next? Two forces pull against each other. What INCREASES it: investment. Households save a share s of their income, and that saving becomes investment (assumption 3): this adds s·f(k) per worker. What REDUCES it: first wear and tear — a share δ of the stock goes to rust and obsolescence each year, i.e. δ·k. Then demography: if the population grows at rate n, the same stock has to be shared among more heads, which lowers capital PER WORKER by about n·k — the capital has not disappeared, it has been diluted. The two add up into a single maintenance requirement, (n + δ)·k. Hence the fundamental equation of the model, the one that was missing: **Δk = s·f(k) − (n + δ)·k**. In words: capital per worker rises by what is invested, minus what it takes to maintain what exists. The whole course consists in reading that line.

Compare the two terms. The first, s·f(k) = s·√k, rises while flattening: that is the concavity of assumption 2. The second, (n + δ)·k, is a straight line through the origin: the maintenance requirement is proportional to the stock. A curve that flattens and a line that rises steadily necessarily cross once — and once only for k > 0. (They also meet at k = 0, but that point does not interest us: it is the economy without a single machine, which produces nothing and can invest nothing. Set it aside and only one crossing is left.) That point, written k*, is the STEADY STATE: there, Δk = 0 and capital stops moving. It remains to show that we get there, and the sign of Δk says so. Take s = 20% and δ = 5% (with n = 0 for simplicity). At k = 4: you invest 0.20 × √4 = 0.40, you must maintain 0.05 × 4 = 0.20 — investment wins, Δk = + 0.20, capital RISES. At k = 25: you invest 0.20 × 5 = 1.00, you must maintain 0.05 × 25 = 1.25 — maintenance wins, Δk = − 0.25, capital FALLS. Below k* we go up, above it we come down: the point is unique, and it is STABLE. Whatever its starting point, the economy lands there.

Let us first solve for k*. Setting Δk = 0 gives s·√k* = δ·k*. We can divide by √k* — non-zero, since k* = 0 is the other solution, that of an economy with nothing — and what is left is s = δ·√k*, so **k* = (s/δ)²** and y* = s/δ. With s = 20% and δ = 5%: k* = 4² = 16 and y* = 4. Double the saving, s = 40%: k* = 64, y* = 8. Capital quadruples, output doubles: more saving does give a higher level. But what do you CONSUME? Not y*, only what is left once maintenance is paid: c* = y* − δ·k*. At s = 20%: 4 − 0.8 = 3.2. At s = 50%: k* = 100, y* = 10, maintenance 5, so c* = 5.0. At s = 70%: k* = 196, y* = 14, maintenance 9.8, so c* = 4.2 — less than at 50%, for a clearly higher output. Three points are not a proof, so let us prove it in two lines, with no derivative at all. Substitute: c* = y* − δ·k* = s/δ − δ·(s/δ)² = (s − s²)/δ. Recognise a parabola in s. Complete the square in the numerator: s − s² = − (s² − s) = − (s − ½)² + ¼. So **c* = [¼ − (s − ½)²] / δ**. Since the square is always positive or zero, something is always taken away from ¼ — except at one point: s = ½, where the square vanishes. The maximum is therefore reached exactly at **s = ½ = α**, and it is worth 1/(4δ) = 5. This is the GOLDEN RULE, and you can see at once why it is symmetric: saving 30% or 70%, twenty points either side, gives strictly the same consumption. Beyond it, the country works for its machines. Click each term: