Using maths how could I show this, I am able to show that AC>MC By differentiating AC with respect to q and assuming AC is falling , but how do show its falling in the first place if we have increasing returns to scale. Thanks

aso I know that IRTS means f(tz)>tf(z) where f is productions function gives us maximum amount of output for a given set of inputs Zi.

  • $\begingroup$ Are you familiar with the cost function? Let $c(q)$ be the cost of producing quantity $q$ (when choosing inputs $\mathbf{z}$ to minimize cost). Can you show increasing returns to scale implies $c(tq) < t c(q)$ for $t> 1$? Then recall that average cost is $\frac{c(q)}{q}$. (Note: increasing returns to scale means $f(t\mathbf{z}) > tf(\mathbf{z})$ for $t>1$.) $\endgroup$ – Matthew Gunn Nov 6 '18 at 17:45

A production function $f: \mathbb{R}^k \rightarrow \mathbb{R}$ exhibits increasing returns to scale if for $\alpha > 1$:

$$ f(\alpha \mathbf{x}) > \alpha f(\mathbf{x})$$ This implies the directional derivative of $f$ at $\mathbf{x}$ in the direction of $\mathbf{x}$ is greater than $f(\mathbf{x})$. (Let $\alpha = (1+ \epsilon)$ then $\frac{f(\mathbf{x} + \epsilon \mathbf{x}) - f(\mathbf{x})}{\epsilon} > f(\mathbf{x})$ then take limit as $\epsilon \rightarrow 0$ to obtain $\nabla_\mathbf{x} f(\mathbf{x}) > f(\mathbf{x})$.)

Given a price vector $\mathbf{p}$ the cost function gives the minimum cost to produce a quantity $q$.

$$ c(q) = \min_\mathbf{x} \left\{\mathbf{p} \cdot \mathbf{x} \mid f(\mathbf{x}) \geq q \right\} $$

The Lagrangian for this problem is $\mathcal{L} = \mathbf{p} \cdot \mathbf{x} + \lambda (q - f(\mathbf{x}))$. The first order condition is $\mathbf{p} = \lambda \nabla f(\mathbf{x}^*))$. Take the dot product of both sides with $\mathbf{x}^*$ to obtain $ c(q) = \mathbf{p} \cdot \mathbf{x}^* = \lambda \nabla f(\mathbf{x}^*) \cdot \mathbf{x}^*=\lambda \nabla_{\mathbf{x}^*}f(\mathbf{x}^*)$.

Our previous result on increasing returns to scale then implies $c(q) > \lambda f(\mathbf{x}^*) = \lambda q$. By the envelope theorem, marginal cost is $\frac{dc}{dq} = \frac{\partial \mathcal{L}}{\partial q} = \lambda$. Combining, this gives us marginal cost is less than average cost for an increasing return to scale production function: $$ \frac{dc}{dq} < \frac{c(q)}{q}$$

That immediately implies average cost is decreasing in $q$.


A more intuitive way to think about this is that increasing returns to scale in the production function implies that for $\alpha > 1$:

$$ c(\alpha q) \leq \alpha c(q)$$

If the production $f$ scales more than linearly in inputs, the cost of producing a quantity scales less than linearly. You can show this without any calculus at all.

Once you have that, divide by $\alpha q$ and you immediately obtain:

$$ \frac{c(\alpha q)}{\alpha q} \leq \frac{c(q)}{q} \text{ for } \alpha > 1 \text{ and IRTS}$$ That shows for an increasing returns to scale production function, a higher level of production has a lower average cost.


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