Does returns to scale always lead to economies of scale? I can tell the difference between returns to scale and economies of scale, but then I still don't know if returns to scale ALWAYS leads to economies of scale?
Please could you help me?
Excellent question (I am assuming your intended question is &quot;does increasing returns to scale always leads to economies of scale&quot;):
The two concepts are related but Returns to scale (RS) is much restrictive than the Economies of scale (ES).
The concept of RS is embedded in production function. If $Q=F(K,L)$ then increasing returns to scale simple means:
$$F(\alpha K, \alpha L) &gt; \alpha F(K,L)$$
For example, in Cobb-Douglas production function: $Q=AK^aL^b$, we have that
$a+b&gt;1 \implies \text{increasing }RS$
The concept of ES is much, much wider and goes beyond production function. All it says is that average cost (AC) decreases with $Q$:
$$\frac{dAC}{dQ}&lt;0$$
Notice the use of $d/dQ$ instead of $\partial/\partial Q$. This is where everything changes. In partial derivative we are interested in the mathematical relationship that: ceteris paribus whether changes in quantity produced changes cost.
To illustrate the relationship and the difference consider derivation of $C=f(Q)$ using the Cobb-Douglas production function:
Given, wage rates $w$ and cost of capital $r$:
\begin{align}
C=wL+rK
\end{align}
To first solve for output maximization constraint by cost function:
\begin{align} 
\max_{L,K} \;{AK^aL^b} \;\; s.t \;\; wL+rK=\bar{C}
\end{align}
Solving the langrangian would give us:
\begin{align}
K=\frac{bw}{ar}L \tag{2}
\end{align}
Substituting $(2)$ in production fuction gives us:
\begin{align}
Q=A\bigg(\frac{bw}{ar} \bigg)^bL^{a+b} \tag{3}
\end{align}
Re-arranging $(3)$ and getting $L$ in terms of $Q$ and then substituting it back to $(2)$ gives us:
\begin{align}
L=\bigg(\frac{ar}{bw}\bigg)^{b/(a+b)} \bigg(\frac{Q}{A}\bigg)^{1/(a+b)} \tag{4}
\end{align}
\begin{align}
K=\bigg(\frac{bw}{ar}\bigg)^{a/(a+b)} \bigg(\frac{Q}{A}\bigg)^{1/(a+b)} \tag{5}
\end{align}
Substituting $(4), (5)$ in $(1)$:
\begin{align}
C=\eta \cdot w^{\frac{a}{a+b}} \cdot r^{\frac{b}{a+b}} \cdot Q^{\frac{1}{a+b}}
\end{align}
where $\eta$ is a constant in terms of $a$ and $b$.
For average cost:
\begin{align}
\frac{C}{Q}=\eta \cdot w^{\frac{a}{a+b}} \cdot r^{\frac{b}{a+b}} \cdot Q^{\frac{1-(a+b)}{a+b}} \tag{6}
\end{align}
Now you see if there are increasing returns to scale, i.e., $a+b&gt;1$,
$$\frac{\partial AC}{\partial Q}&lt;0$$
So you see, if $w,r$ can be taken as constant then of course
$$\frac{\partial AC}{\partial Q} = \frac{dAC}{d Q}  &lt;0$$
On the other hand, this is rarely true. In a general equilibrium model, $w$ and $r$ are also variables. For instance, say the Labor and Capital market is perfectly competitive:
$w=MP_L\equiv \frac{\partial Q}{\partial L} = a\frac{Q}{L}$ and similarly, $r=MP_K\equiv \frac{\partial Q}{\partial K} = b\frac{Q}{K}$
Substituting these in $(1)$ (or equivalently in $(6)$), we get:
$$\frac{C}{Q}=(a+b)$$
So interestingly, the average cost is constant despite having increasing returns to scale.
So, increasing RS can at best ensure:
$$\frac{\partial AC}{\partial Q}&lt;0$$
But what ES requires is:
$$\frac{dAC}{dQ} = \frac{\partial AC}{\partial w}\frac{\partial w}{\partial Q}+\frac{\partial AC}{\partial r}\frac{\partial r}{\partial Q}+\frac{\partial AC}{\partial Q}&lt;0$$
So it is perfectly possible that $\frac{\partial AC}{\partial Q}&lt;0$ but $\frac{d AC}{d Q}&gt;0$