<p>A production function <span class="math-container">$f: \mathbb{R}^k \rightarrow \mathbb{R}$</span> exhibits <em>increasing returns to scale</em> if for <span class="math-container">$\alpha &gt; 1$</span>:</p>

<p><span class="math-container">$$ f(\alpha \mathbf{x}) &gt; \alpha f(\mathbf{x})$$</span>
This implies the directional derivative of <span class="math-container">$f$</span> at <span class="math-container">$\mathbf{x}$</span> in the direction of <span class="math-container">$\mathbf{x}$</span> is greater than <span class="math-container">$f(\mathbf{x})$</span>. (Let <span class="math-container">$\alpha = (1+ \epsilon)$</span> then <span class="math-container">$\frac{f(\mathbf{x} + \epsilon \mathbf{x}) - f(\mathbf{x})}{\epsilon} &gt; f(\mathbf{x})$</span> then take limit as <span class="math-container">$\epsilon \rightarrow 0$</span> to obtain <span class="math-container">$\nabla_\mathbf{x} f(\mathbf{x}) &gt; f(\mathbf{x})$</span>.)</p>

<p>Given a price vector <span class="math-container">$\mathbf{p}$</span> the cost function gives the minimum cost to produce a quantity <span class="math-container">$q$</span>.</p>

<p><span class="math-container">$$ c(q) = \min_\mathbf{x} \left\{\mathbf{p} \cdot \mathbf{x}  \mid f(\mathbf{x}) \geq q \right\} $$</span></p>

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

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

<p>That immediately implies average cost is decreasing in <span class="math-container">$q$</span>.</p>

<h3>Intuition</h3>

<p>A more intuitive way to think about this is that increasing returns to scale in the production function implies that for <span class="math-container">$\alpha &gt; 1$</span>:</p>

<p><span class="math-container">$$ c(\alpha q) \leq \alpha c(q)$$</span></p>

<p>If the production <span class="math-container">$f$</span> 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.</p>

<p>Once you have that, divide by <span class="math-container">$\alpha q$</span> and you immediately obtain:</p>

<p><span class="math-container">$$ \frac{c(\alpha q)}{\alpha q} \leq \frac{c(q)}{q} \text{ for } \alpha &gt; 1 \text{ and IRTS}$$</span>
That shows for an increasing returns to scale production function, a higher level of production has a lower average cost.</p>

CC BY-SA 4.0
2018-11-07T00:03:34.610Z
2018-11-07T04:05:23.347Z
2018-11-07T04:05:23.347Z
Matthew Gunn
<p>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</p>

<p>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.</p>

CC BY-SA 4.0
2018-11-06T17:45:02.087Z
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 $t>1$.)
Matthew Gunn
<microeconomics>
How to show average cost is falling if we have IRTS
Rivaldo
# How to show average cost is falling if we have IRTS

(Asked by: username_0 on 2018-11-05)

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.

**Tags:** microeconomics

> **Comments:**
> **username_1**: 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 $t>1$.)  

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## 1 answers

### Answer by username_1 on 2018-11-07. Score: 1 | Accepted answer

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.


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