I learned that when there is decreasing returns to scale, the average cost is always increasing.

But the professor told us today that the other way around might not always be true. So if average cost is increasing, it might not necessarily mean that there is decreasing returns to scale.

The production function $q = \min \{K, L^{(1/2)}\}$ was one he gave us as something to think about as a possible counterexample.

we assumed that the input prices are 2 for capital and 1 for labor.

If we solve for this using cost minimization, we get $C(q) = 2q + q^2$ and the $AC(q) = 2 + q$ which is an increasing function.

And if I input $kK, kL$ in the production function where $k > 1$ , then the function becomes

$q = \min \{kK, (kL)^{1/2}\} = \min \{kK, k^{1/2} * L^{1/2}\}.$

How do I know what returns to scale this is showing?

if the min function was simply $q = \min \{ K, L \}$, then I could definitely tell that this is showing constant returns to scale since $\min \{kK, kL\} = k*\min\{K,L\}$, but how do you mathematically solve for returns to scale for a function like the one above, $q = \min \{K, L^{1/2}\}$ ?

I don't know what step to take next after arriving at $\min \{kK, k^{1/2} * L^{1/2}\}$. I don't know what to extract out of the min function from this point on.

I would appreciate advice on how to continue to determine the returns to scale for this kind of example.

  • $\begingroup$ How did you define constant/increasing/decreasing returns to scale in your course? $\endgroup$ Dec 17, 2020 at 9:53
  • $\begingroup$ en.wikipedia.org/wiki/Returns_to_scale just like the wikipedia (Formal definitions part ) so I need to find the relationship between f(kK, kL) and kf(K,L), and the function of interest is a bit difficult for me to derive it... $\endgroup$
    – Robin311
    Dec 17, 2020 at 10:19
  • $\begingroup$ If this has to hold for all $K$ and $L$, then neither of these categories apply. $\endgroup$ Dec 17, 2020 at 10:25
  • $\begingroup$ could you explain what you mean in detail? Do you mean that it's not IRS, DRS, or CRS? But shouldn't a production function show one of the three for any time you multiply k ? $\endgroup$
    – Robin311
    Dec 17, 2020 at 11:19
  • $\begingroup$ For example, increasing returns to scale means that for all $K$, all $L$, and all $\lambda>0$ one has $f(\lambda K,\lambda L)>\lambda f(K,L)$. This is violated as soon as there exists at least one $K$, $L$, and $\lambda>0$ such that the inequality holds. But that does not guarantee that the converse inequality or equality hold in every case. $\endgroup$ Dec 17, 2020 at 14:17

2 Answers 2


You have a Leontief production function and in optimum you will always have $K=\sqrt{L}=q_1$. Now increase both inputs by factor $k>1$ and you arrive at $k*K > \sqrt{kL}=q_2$ where the first inequality follows from $k>\sqrt{k}$ for $k>1$ and the second ineqality from the fact that only the minimum matters in your production function. Therefore you have $F(kK,kL)= \sqrt{kL}<kK=k\sqrt{L} =k*F(K,L)$ with any optimal input combination $(K,L)$ and therefore decreasing returns to scale.

  • $\begingroup$ I thought returns to scale needed to satisfy for any K and L, not just at the optimum..... $\endgroup$
    – Robin311
    Dec 17, 2020 at 12:37
  • $\begingroup$ Here, this is not true as a comment already suggested. Take $K=1$ and $L=10$ giving output $q_1=1$, then you have CRS for low $k$, e.g. $k=2$ gives you output $q_2=2=2*q_1=2$. $\endgroup$
    – Bayesian
    Dec 17, 2020 at 12:50
  • $\begingroup$ Also note that your cost function also relies on the optimal combination of K and L. $\endgroup$
    – Bayesian
    Dec 17, 2020 at 12:54
  • 1
    $\begingroup$ So are you saying that a counterexample could be like, when K < L^(1/2), F(K,L) = K and if kK < k^(1/2) * L^(1/2), then F(kK, kL) = kK = kF(K,L) which means it's CRS in this case instead of DRS ? $\endgroup$
    – Robin311
    Dec 17, 2020 at 13:01
  • $\begingroup$ Yes, that's what I meant. Given this example, I would be curious how your professor arrives at DRS for general K,L. $\endgroup$
    – Bayesian
    Dec 17, 2020 at 13:23

To understand what is the issue here, try dutifully to examine all possible sub-cases in the production function.

The production function is

$$Q_0 = \min\{K_0, L_0^{1/2}\}$$.

Consider cases

A. $K_0 < L_0^{1/2}$

Here $Q_0 = K_0$. Consider $Q_{\lambda }\equiv \min\{\lambda K_0, \lambda^{1/2}L_0^{1/2}\},\;\;\; \lambda>1.$

Subcase A1.
If $\lambda K_0\leq \lambda^{1/2}L_0^{1/2} \implies Q_{\lambda} = \lambda K_0 = \lambda Q_0$ and we have Constant Returns to Scale. This happens for $K_0 \leq \left(L_0/\lambda\right)^{1/2}$.

Subcase A2.
If $\lambda K_0 > \lambda^{1/2}L_0^{1/2} \implies Q_{\lambda} = \lambda^{1/2}L_0^{1/2}$ and because we are in the case $K_0\leq L_0^{1/2}$ and $\lambda >1$ we have $Q_{\lambda} > Q_0$ and we have Increasing Reaurns to Scale. This happens for $\left(L_0/\lambda\right)^{1/2}< K_0 < L_0^{1/2}$.

B. $K_0 > L_0^{1/2}$

Here $Q_0 = L_0^{1/2}$. Consider $Q_{\lambda }\equiv \min\{\lambda K_0, \lambda^{1/2}L_0^{1/2}\},\;\;\; \lambda>1.$

Here, because $\lambda >1$, we will always have $\lambda K_0 > \lambda^{1/2}L_0^{1/2} \implies Q_{\lambda} = \lambda^{1/2}L_0^{1/2} = \lambda^{1/2} Q_0$ and so we obtain Decreasing Returns to Scale.

C. $K_0 = L_0^{1/2}$

Here too, because $\lambda >1$, we are in the same situation as in Case B, so Decreasing returns to Scale.

So it appears that the "point of departure" (the input situation for $Q_0$) matters -and this is why in the other answer, optimal behavior was invoked, in order to pin down in some meaningful sense this "point of departure".


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