As I like to do before getting down to some hardcore maths, I’m going to take an artistic break and present to you a Haiku that I wrote a couple of days ago:

*The flower withers*

*Petal falls on wilted earth*

*New shoot takes its place*

There, that’s enough of the artsy fartsy stuff for now.

##### What’s all this about?

I will firstly explain briefly what cost curves are, and why they look like they do. I will then go on to prove the fact that the Marginal Cost curve intersects the Average Cost curve at its lowest point. Using some simple but fancy maths, of course.

##### Cost Curves

I’m sure you’ve noticed the 3 curves in the graph above. These generic graphs show us the various costs incurred by a typical firm when they produce some good (e.g a box of chocolates). The horizontal axis represents the quantity (q) of the item produced, and the vertical axis represents the cost (C) of producing q amounts of the good.

Let’s roll with the boxes of chocolate example. The first curve we’re interested in is the **total cost (TC) curve**. It has 2 key features:

1. The curve is always increasing because it obviously costs you more money in total to produce more boxes of chocolate (and also the cost is never negative!)

2. The more boxes of chocolate you produce, the more expensive each additional item becomes to produce. In other words, we can say the marginal cost (MC) is increasing. In the graph, this means that it gets steeper as q increases. (A reason for this could be because the machines used to produce the chocolate need additional maintenance the more chocolate is produced.)

The second point does not always apply, but it makes our example neat, because we can represent the TC curve with a quadratic equation:

**TC = aq ^{2} + bq + c**

In this general function a, b and c are just constants – any 3 fixed numbers.

The **average cost (AC) curve** represents the average cost of producing q boxes of chocolate (Easy isn’t it?). How do you work out the average? By taking the total and dividing it by how many you have:

**AC = (TC/q) = aq + b + c/q**

You can see that the AC curve is sort of ‘U’ shaped – it starts off high, then goes down until it hits a minimum point, and finally it goes up and keeps going up.

The **marginal cost (MC) curve** represents the marginal cost of q – which is the cost of producing each extra box of chocolates. Remember when I said the marginal cost is increasing because the slope of the TC curve is increasing? Well this is because the MC curve is just the slope of the TC curve. We work this out by differentiating the TC function:

**MC = (dTC/dq) = 2aq + b**

This is the equation of a straight line, which explains the graph above. This tells us, for example, that the 3rd box of chocolates is more expensive for you to produce than the 2nd, which is more expensive than the 1st, and so on.

##### The thing we want to prove

Just from looking at the curves, you might notice that the MC curve crosses the AC curve at exactly its minimum point. That is no coincidence my friend, and we are going to prove that it is always true.

I’ll explain why this fact should be true intuitively. When the average cost is falling, it means that the cost of producing each extra good must be less than the average (otherwise it wouldn’t go down). Conversely, when the average is rising, the cost of each extra good must be greater than the average. Only when the cost of the next good is exactly equal to the average, will the average remain constant. And when the average remains constant, it is at the stationary point of the graph (its lowest point in the ‘U’).

To simplify this, take a list of numbers 1,2,3. The average is 2 because (1+2+3)/3 = 6/3 = 2. If the next number we added to the list was 6 (higher than the average) then the average would rise to 12/4 = 3.

If instead we added a 1 to the list, the average would fall to 7/4 = 1.75.

But if we added a 2 to the list (which is exactly the average), then the new average would be the same as the number we added – 8/4 = 2.

##### The Proof

That explanation was nice, but we need to do some maths to show that the MC always crosses that AC at its lowest point.

First, we find the lowest point of the AC curve, by differentiating AC (denoted by AC’) and setting it equal to 0:

**AC’ = a – c/q ^{2} = 0**

so a = c/q^{2} which means that **q = √(c/a)** at the lowest point of AC.

Now we need to show that the point where AC = MC has the same value of q as we have found above.

Let AC = MC. Then:

aq + b + c/q = 2aq + b

aq = c/q

q^{2} = c/a

**q = √(c/a)**

And as if by magic, the point where AC and MC cross is the same as the lowest point of AC. Hence this proves our statement.

*(As a disclaimer, this only strictly proves that the statement is true for quadratic cost functions. But the statement is true for all functions, and this can be proved pretty easily. However, the notation becomes a bit advanced and abstract, which is why I’ve only shown the proof for the neat quadratic case.)*

Great stuff! Thanks!

This only proves that it is true for a linear cost function

I meant for quadratic equation. Full proff is here:

Consider MC(q) = C(q)’ (the derivative of the cost with respect to q)

and AC(q) = C(q) / q.

Then AC(q)’ = [ C(q)’ * q – C(q) ]/ q^2. Set AC(q)’ = 0. hence by multiplying by q^2 we get:

C(q)’ * q – C(q) = 0 (1)

On the other hand. If we set MC(q) = AC(q) we have

C(q)’ = C(q)/q. Multiplying by q then subtracting C(q) we get

C(q)’ * q – C(q) = 0 which is indeed the same equation as (1). Hence we find that solving the system to find the minimum of AC(q) si equivalent to finding the intersection of MC(q) with AC(q) which proves that these are in fact the same problem.

Correct, hence my disclaimer at the end. This was designed to be an intuitive introduction (note the tone of the article).