Marginal Cost is $$MC(y) = \frac{\partial C(y)}{ \partial y}$$
Average Cost is $$AC(y) = \frac{ C(y)}{ y}$$ Average Variable Cost is $$AVC(y) = \frac{VC(y)}{ y}$$ Note: Cost is $$ C(y) = FC + VC(y)$$ I don’t understand what the differences are. Like I understand the definition of cost function, fixed costs, and variable costs, but I don’t get how marginal cost, average cost, and average variable costs are related.

Can someone explain?

  • $\begingroup$ @EnergyNumbers voted to close but this one is actually slightly more complete (includes AVC). Given the author is the same, will close nevertheless. $\endgroup$
    – VicAche
    Commented Dec 3, 2015 at 13:01

2 Answers 2


Suppose initially there are no fixed costs.

What does it mean to take an average?

Consider a cost function $C(y)$. What does it mean to take the “average” of this function? Mathematically, it is just $$A(y) = \frac{c(y)}{y}$$

Let’s suppose we are considering $C(y) = y^3$. Suppose we now consider $y = 5$. Then $$A(5) = \frac{5^3}{5} = 25 $$ This is just saying that, for each unit I buy, I am buying them at $25$ each on average. So I could have paid $$ \frac{15 + 39 + 46 + 14 + 11}{5}$$ or $$ \frac{17 + 3 + 78 + 23 + 4}{5}$$ What information does the average cost give us?

The 'cost per unit' given by an average cost function isn't like taking an average by adding up the cost of each unit we bought. When we are given a cost function $C(y)$, this just tells me the total cost. I don't know how much my first TV cost me from this equation alone. And the average doesn't tell me that either. Note how above we have two sets of 5 TVs that yielded the same average cost. Thus, the average cost function doesn't tell me how much I paid for each specific TV.

What does marginal cost tell me?

This is exactly what marginal cost provides. Marginal cost provides the specific cost of each successive infinitesimal amount of good. Consider again $C(y) = y^3$. $$MC(y) = \frac{d C(y)}{d y}= 3y^2$$

Suppose I have purchased $3.5$ units of TV. Then the MC equation thus says, for $c(y) = y^3$, each additional infinitesimal amount of TV costs me $$3(3.5)^2 = 36.75$$ at that point. If I purchase an additional $0.1$ amount of TV, then my MC changes and now I am at $$3(3.6)^2 = 38.88$$

Thus, this concept is a bit trickier because it involves a continuous amount of TV and the cost per infinitesimal unit changes as you buy more. So you really can't just consider how much $MC(1)$ is and $MC(2)$ is. You are considering it for some infinitesimal additional amount $dy$ at a given point (e.g. $y=2$). The trend is easier to think through if you don't restrict yourself to integers. Note, we are assuming you can have noninteger quantities of goods, otherwise we wouldn't be working in $\Bbb{R}^n$ and the integration later might be trickier.

Example to Clarify Marginal Cost

So suppose I want to buy $5$ TVs. For the $k$th TV purchased, the $$MC(k)= 3k^2$$ Going back to our example above, let's now suppose $k \in [0,5]$. To find the average cost, we simply do the addition formula used above for the 5 TV example, except now summed over each infinitesimal amount (of which there are an infinite number). This gives us

$$A(5) = \frac{3(0)^2+\cdots + 3k^2+ \cdots + 3(5)^2}{5} = \frac{\int_{0}^{5} 3y^2 dy}{5}= \frac{(125-0)}{5}=\frac{c(5)}{5}=25$$

$$A(5) =25$$ the same as we calculated earlier.


Marginal Cost is $$MC(y) = \frac{d C(y)}{d y}$$ Average variable cost is

$$A(y) = \frac{\int_{0}^{y} MC(y) dy}{y}$$

Note, since we assumed $FC = 0$, this formula also thus defines AC but would not be true for $FC \neq 0$.


I have been sloppy about the distinction between AVC and AC up until now. I have avoided this distinction by assuming $FC = 0$. Now I will try to clarify this point by assuming $FC$ can be anything.

Fixed costs ($FC$)are costs the firm pays that do not vary with the amount produced. For example, suppose I work for Uber and I buy a car. That money is spent and doesn't change with the amount I drive. But the amount of gasoline I consume does change as with the amount I drive. Costs that scale with production are known as variable costs ($VC(y)$).

$$AC(y) = \frac{C(y)}{y}=\frac{VC(y) +FC}{y}= \frac{VC(y)}{y} + \frac{FC}{y}=AVC + AFC $$

Hyperbolic Behavior of AFC with y >0

As production goes up ($y\rightarrow \infty$), AFC goes down ($AFC \rightarrow 0$) in an inversely proportional fashion. But note, as production goes down ($y\rightarrow 0$), AFC goes to infinity ($AFC \rightarrow \infty$). Thus, the plot of AFC will always be a hyperbola unless $FC = 0$ in which case AFC is just 0.

How does AFC affect AC?

Without specifying $VC(y)$, we cannot know the behavior of $AC(y)$ as $y$ moves away from $0$. There will be some $y_{0}$ such that, for $y > y_{0}$, $AC(y)$ can essentially be anything. Of course, as $y \rightarrow 0$, $AC(y) \rightarrow \infty$. This is because AVC cannot be negative and so we are guaranteed any variable costs will not lower the average cost below AFC. Therefore, $AVC \geq 0$, so since $AFC \rightarrow \infty$ as $y\rightarrow 0$ (remember it is a hyperbola), thus AC must go to infinity as well.

Thus, since the behavior of fixed costs are always known, AVC is the missing ingredient needed to specify the behavior of AC.

AC and AVC can't be exactly equal for $FC>0$

Since $AFC \rightarrow 0$ but does not ever equal 0 (for $FC >0$), we know that $$AC \neq AVC$$ for any $y$ and $FC >0$. But since $AFC$ approaches 0 for large enough $y$, AC approaches AVC asymptotically.

AC and AVC can't be exactly parallel for $FC>0$

If $AVC$ and $AC$ are parallel, then their derivatives should be equal. But notice that $$\frac{dAFC}{dy} = -\frac{FC}{y^2}$$ so $$\frac{dAC}{dy} = \frac{dAVC}{dy} + \frac{dAFC}{dy} = \frac{dAVC}{dy} - \frac{FC}{y^2} \neq \frac{dAVC}{dy}$$ So they aren't parallel because their derivatives aren't equal. That said, for large enough $y$, the derivatives will be very close to each other so they may appear nearly parallel over some portion of the curves.

MC intersects AC at minimum point of AC curve

See Alecos's answer.

MC intersects AVC at minimum point of AVC curve

Consider the average variable cost curve. Find $y^*$ that solves

$$\min_{y} AVC(y)$$

So at this point, $$\frac{dAVC(y^*)}{dy^*} = 0$$

This means by quotient rule

$$\frac{VC'(y^*)y^*-VC(y^*)}{(y^*)^2} = 0$$ and since $y^*$ can't be 0, this implies $$VC'(y^*)y^*-VC(y^*)=0$$ which rearranged gives $$VC'(y^*)=\frac{VC(y^*)}{y^*}$$ Recall, $$C(y) = FC + VC(y)$$ Note, since $FC$ is a constant, $$C'(y) = VC'(y)$$ Therefore,

$$C'(y^*) = \frac{VC(y^*)}{y^*}$$


Why do we define the firm's longrun shutdown point in terms of average cost not marginal cost?

Recall a firm's profit function is $$\pi = py - C(y)$$

I am not defining the behavior of $p$ here or who controls $p$. I am just considering for what $p$ will the firm shut down, and ignoring everything else because it's irrelevant.

So we can easily see that since $$AC(y)= \frac{C(y)}{y}$$ for $p=AC(y)$, this yields $$\pi = \left(\frac{C(y)}{y}\right)y - C(y) = C(y)- C(y) = 0$$ So the firm will shut down if $p <AC(y)$ because then $\pi < 0$.

So consider the function $C(y) = y^3$. Note, $$MC = 3y^2 > AC = y^2$$ We know the firm profit maximizes at $MR = MC$. So since, for $y > 0$, $MC(y) > AC(y)$, the firm would always produce if $p = MC$ since this would mean for $y>0$,

$$ MC(y)y-C(y) = \pi_{p=MC} > \pi_{p=AC} = \left(\frac{C(y)}{y}\right)y - C(y) =0$$
$$\pi_{p=MC} > \pi_{p=AC} = 0$$ So, for this $C(y)$, at $p=MC$, $\pi >0$ for all $y$.

So this example clearly shows you would never shut down at $p=MC$ for $y>0$ for $C(y) = y^3$. Although this isn't the most thorough explanation, this example clearly invalidates that train of thought and makes clear firms shut down for $p< AC$

Does average cost necessarily equal marginal cost at some point?

See Alecos's answer.

Takeaway Rules

  1. In long run, firms produce if $p \geq AC(y)$. They shut down for $p < AC(y)$
  2. Firms always produce at $MR = MC$
  • $\begingroup$ While technically it is true that all firms produce at $MR = MC$, the way you formulated it makes it seem like all firms are monopolies. $\endgroup$
    – Giskard
    Commented Dec 3, 2015 at 7:22
  • $\begingroup$ @denesp Did I? How would you change it? You can propose an edit and I am happy to accept. I'm sure you know of better notation than I used. $\endgroup$ Commented Dec 3, 2015 at 8:22
  • $\begingroup$ I have qualms about this part: "Recall a firm's profit function is $\pi = p(y)y - C(y)$". In a competitive market this usually takes the form $\pi = p \cdot y - C(y)$ instead. I am afraid I have no suggetion for an edit as the answer is very long and I do not have a full grasp of it. $\endgroup$
    – Giskard
    Commented Dec 3, 2015 at 11:06
  • $\begingroup$ (+1) for the effort (although @denesp comments stand), but I spotted a mistake in your answer. Check my post. $\endgroup$ Commented Dec 3, 2015 at 14:03

Does average cost necessarily equal marginal cost at some point?

Yes, if the Average Cost function has an interior stationary point.

Considering the derivative of the Average Cost function with respect to quantity, we have

$$\frac {\partial AC}{\partial y} = \frac {\partial (C/y)}{\partial y} =\frac {MC\cdot y - C}{y^2} = \frac {MC - AC}{y}$$


$$\frac {\partial AC}{\partial y} = 0 \implies MC = AC$$.

If the Average Cost function has such a stationary point (i.e. where its first derivative equals zero), will it be a minimum?

Denoting $C''$ the 2nd derivative of the total cost function, we need

$$\frac {\partial^2 AC}{\partial y^2} |_{AC=MC} >0 \implies \big[C''-\big(\partial AC/\partial y\big)\big]\cdot y - (MC-AC) >0$$

and since we aveluate at $y: AC = MC, \;\; \partial AC/\partial y =0$ we get

$$\frac {\partial^2 AC}{\partial y^2} |_{AC=MC} >0 \implies C''> 0$$

The Total Cost Function must be convex at the stationary point, for it to represent a minimum for the Average Cost function.

The Total Cost function used in the OP's example is $TC = 50 + y^3$.

The Average Cost function is $AC = (50/y) + y^2$, and the Marginal Cost is $MC = 3y^2$. So

$$AC = MC \implies (50/y) + y^2 = 3y^2 \implies 25 = y^3 \implies y^* = 25^{1/3} \approx 2.92$$

At that point, the second derivative of the Total Cost function is $C''(y^*) = 6y^* >0$. So for this Cost function there is a point where Average Cost is minimum, and where it equals Marginal Cost. The Diagram is

enter image description here

Indirectly, the thing to remember here is that the nice $J$-shaped curve for the Average Cost function is most easily produced if we allow for fixed costs and a convex Variable Cost.


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