# How to show average cost is falling if we have IRTS

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.

• 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 $t>1$.) 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 $$q$$.

### Intuition

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.