Cobb-Douglas production function, given $w$ get $r$ regardless of input levels. Why?

There is a market economy with technology given by:

$$Y = K^\alpha L^{1-\alpha} \tag{1}$$

Firms behave competitively and input prices are:

$$r = \alpha K^{\alpha-1}L^{1-\alpha} = \alpha(\frac{L}{K})^{1-\alpha} \tag{2}$$

$$w = (1-\alpha) K^\alpha L^{-\alpha} = (1-\alpha) (\frac{L}{K})^{-\alpha} \tag{3}$$

I can solve for $$\frac{L}{K}$$, and obtain the following expression:

$$r = \alpha (1-\alpha)^{\frac{1-\alpha}{\alpha}}w^{\frac{\alpha-1}{\alpha}} \tag{4}$$

Which is a function that links the prices independently of the inputs combination.

1) Is this correct?

2) What is the economic interpretation of the last result?

A clarification: The values $$w,r$$ are not independent of input combinations: you have already solved for $$K/L$$.

1) Unless you assume that the price of the output is equal to 1, there is a minor mistake, as output prices affect factor prices.

2) If there is a profit maximizing pair $$(K,L)$$, then for all $$\alpha \in \mathbb{R}_+$$ $$(\alpha K,\alpha L)$$ will also be profit maximizing. This is because the production function you are looking at has constant returns to scale (CRS).

A consequence of CRS is that if the function is differentiable then both marginal products will be homogeneous of degree zero, hence $$r$$ and $$w$$ will only determine the ratios of $$K$$ and $$L$$, not their level.

• Thanks, @denesp. What I find interesting is that, for instance, given $r$, I can get the value of $w$ that allows for a profit-maximizing plan to exist, independently of $\frac{K}{L}$. This can be useful, I guess. when we compute the steady-state allocation in an economy with CRS Cobb-Douglas; the interest rate is fixed by the consumption side and, thanks to that equation, we can easily get the wage. Do you think this argument is correct? – Tecon Jan 25 '19 at 9:31
• "given $r$, I can get the value of $w$ that allows for a profit-maximizing plan to exist" This is true. "independently of $\frac{K}{L}$". Again, I don't know what you mean by this. $K$ and $L$ are not parameters, they are endogenous variables and will take values as functions of $r$ and $w$. – Giskard Jan 25 '19 at 14:03
• On the macro argument: Sure, but I have always been wary of the CRS assumption in macroeconomics models. – Giskard Jan 25 '19 at 14:03

This is a subtle issue. First let's present a numerical example to see the head-scratching riddle. Assume $$\alpha =1/2 \implies Y = K^{1/2}L^{1/2}$$

and that the exogenously given input prices are

$$r=1/8,\,\, w=4.$$

The f.o.c are

$$\begin{cases} \frac {Y}{2K} = 1/8 \\ \\ \frac{Y}{2L} = 4 \end{cases} \implies K/4 = 8L \implies \left(L/K\right)^* = 1/32$$

This gives us the value of $$L/K$$ that satisfies both first-order conditions.

But let's follow the OP's logic, and write/solve the f.o.c. as

$$\begin{cases} \frac {K^{1/2}L^{1/2}}{2K} = 1/8 \\ \\ \frac{K^{1/2}L^{1/2}}{2L} = 4 \end{cases} \implies \begin{cases} \frac {L}{K} = 1/16 \\ \\ \frac{L}{K} = 1/64 \end{cases}$$

Oops. The system now seems impossible. But it appears that just above we have solved the exact same system. And we didn't do any illegal math operation, like dividing by zero or anything so how is it possible to obtain the above contradictory results?...

But indeed we did something "illegal", although not in terms of mathematical operations in the narrow sense: by solving each f.o.c. separately with respect to $$L/K$$ we transformed a problem of optimizing a bivariate function that leads to a system with two unknowns and two equations, into a problem of directly solving a system with one unknown ($$L/K$$), and two equations.
No surprise then that it will have a solution only for specific combinations of the exogenous parameters, which is expressed by eq. $$(4)$$ of the OP. But this is no longer the profit maximizing problem of the competitive firm.

Another way to say it is that, given that what is treated as exogenous here is the input prices, not all input ratios satisfy the f.o.c.

• Thanks for your answer. I don't understand this sentence: "What this relation says is that if we fix a labor/capital ratio, and we want production to be carried out with this ratio, then the combinations (r,w) must obey relation (4)". Why do you say "fix the ratio"? I solve for the ratio, hence, for any given ratio, that relationship has to be satisfied. And yes, all price combinations are permissible, however, only the price combinations governed by that relationship allow for a profit-maximizing plan to exist, i.e. there is an interior solution of the PMP. – Tecon Jan 25 '19 at 9:17
• @Tecon In response to your comment I completely re-worked my answer. – Alecos Papadopoulos Jan 25 '19 at 11:57
• Thank you very much @Alecos. Your answer really helped me. – Tecon Jan 25 '19 at 14:37
• Alecos, if you are interested, I posted a follow-up question here – Tecon Jan 25 '19 at 16:18
• I am not sure anymore about the correctness of this sentence: "But this is no longer the profit-maximizing problem of the competitive firm." The solution to the PMP of the firm is determined by a system of two equations. This system is "degenerate" in the sense that it can be expressed by a combination of the two unknowns, the ratio $\frac{L}{K}$. Both equation need be satisfied, and this does not happen for any combination of prices, but only for those combinations that satisfy equation (4). – Tecon Jan 29 '19 at 9:41

The system of FOC of the profit maximization problem when the production function is constant return to scale is a degenerate system. For a good explanation see this. In short, it means that the system of two equation in two unknown is actually a system of two equation in one unknown, which in this case is $$\frac{K}{L}$$.

The solution to this problem needs to satisfy both equations (2) and (3), and this happens only if equation (4) is satisfied.

Hence, to answer the main question, given one price, the other is uniquely identified by equation (4). Given one price, to determine the other it is enough to know the technological parameter, in this case, this is just $$\alpha$$.