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I'm really confused about how I'd tackle this.

I have an equation, $TC(Q) = 100Q + 20Q^2 + 3Q^3$.

And I'm trying to find where the economies of scale, diseconomies of scale and constant return to scale occurs.

My instinct would be to take the derivative of that function, and then set it to 0, and then see what the values to the left and right of 0 are which would be the slope and allow me to see depending on whether or not they're positive or negative whether it's an economy or diseconomy of scale.

Is this along the right road?

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To tackle this problem, I'd start by looking up the technical definition of 'economies of scale'. A google search turns up a wikipedia page which says it refers to the case in which average costs are decreasing. To me, that suggests it might be a good idea to start by computing average costs, not the derivative of costs. –  NickJ May 26 at 18:38

1 Answer 1

This is easier when one has it formulated as $Q(TC)$. Here, we have $TC(Q)$. Intuitively, constant returns to scale (CRS) require that costs change one-for-one with quantity. However, this does not translate easily into an expression using $TC(Q)$ only. As @NickJ hinted at, it's easier to work with average costs.

Using average costs In other words, CRS implies that average costs do not change. One way to look for that is to say that we have increasing (decreasing) returns to scale whenever $\frac{d \frac{TC(Q)}{Q}}{dQ}$ is negative (positive).

Using total costs Let's start with the previous expression and see whether we can get a direct formula using $TC(Q)$.

$$\frac{d \frac{TC(Q)}{Q}}{dQ} = (TC'(Q)Q-TC(Q))/Q^2 = 0 \\ TC'(Q)Q - TC(Q) = 0 \\ \frac{TC(Q)}{Q} = TC'(Q)$$

It's not as simple as I initially thought, but we get a meaningful expression: We have CRS whenever average costs are equal to the marginal change in total costs.

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FooBar, I'm not sure I understand your answer. Are you saying that a firm with $TC(Q) \equiv 100$ everywhere has constant returns to scale everywhere? I would say intuitively that such a firm has decreasing returns to scale. –  NickJ May 27 at 3:28
@NickJ In my defense, it was late. I was in the mindset of $Q(TC)$ instead of $TC(Q)$. –  FooBar May 27 at 13:46

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