# Why does negative derivative of AC with respect to q imply Economies of scale (Monopoly)

In the following Industrial organization exercise form Church and Ware chapter 4 exercise 2

(preliminary info: C(q) = cost function; f = fixed costs; c = marginal costs; q= quantity of goods produced)

2.) Suppose that the cost function for both the entrant and an incumbent is $$C(q) = f + cq$$

(a) Is this technology characterized by economies of scale? What is marginal cost and how does it compare to average cost?

(b) Suppose that postentry the incumbent can commit to charge p = c. Will there be entry? Does it matter whether f is sunk or not?*

my professor's solution for question (a) was just stating "since the first derivative of AC is negative $$-\frac{f}{q^2}$$ the technology is characterised by economies of scale"

Could anyone argument on that? I got to the same conclusion by just looking at average cost function $$AC=\frac{f}q+c$$, where as quantity grows, fixed cost diminish until the whole function becomes almost flat and ≈ c

What I tried to do:

I graphed the derivative, it's a parabola facing upward that goes to plus infinity around y = 0 and plus infinity around x = 0 (we are only interested in the first quadrant values since P and Q negative don't make sense) this is because f is a negative number (being a cost). But still I don't understand the relevance of the sign, what I feel is relevant is the flattening of the function as q grows, this is what determines economies of scale for what I know.

Thanks to anyone helping out.

Could anyone argument on that?

Your professor was correct. For economies of scale to be present the $$AC'_q<0$$. This is because $$AC'_q<0$$ says that average costs fall when you produce more output which by definition are the economies of scale (economies of scale mean that production gets cheaper as you produce more and more output).

I got to the same conclusion by just looking at average cost function $$AC=\frac{f}{q}+c$$, where as quantity grows, fixed cost diminish until the whole function becomes almost flat and $$≈ c$$

You can usually solve most problems in more than one way. For example, maximizing profit function can be done analytically by calculus (e.g. taking derivative of the profit function wrt choice variables, finding stationary points, checking second order conditions), but the same problem can be solved graphically by plotting the function and visually locating highest point, or it could be solved numerically (for example by the Newton's method).

The same way you can figure out that there are economies of scale present by various methods. You could for example do it visually by graphing $$AC$$ function.

• Thank you for the answer, but shouldn't having q at denominator in AC function be enough then? He took the derivative as a further step and only considering that as a solution, that's what put me off. Aug 8, 2022 at 10:02
• @AdrianoPollio enough for what? If you want to prove it rigorously using calculus then its not enough to just reason from knowledge of shape of functions. Again there are multiple ways how to solve this. Also, your professor might require you to learn the calculus method because this example might be easy but for a very complex mathematical function it might be impossible to eyeball it
– 1muflon1
Aug 8, 2022 at 10:08
• I just realized both you and my professor were using derivative to check the primitive shape (negative first derivative implying decreasing primitive function). That's the thing I was missing and I focused on the derivative itself without considering the relation with the primitive (you gave me a hint as you suggested that my professor wanted to use calculus, unfortunately his slides are very poor and also lectures). Thanks again Aug 8, 2022 at 10:15