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Marginal cost is the cost associated with producing one more unit of output. Mathematically speaking, marginal cost is equal to the change in total cost divided by the change in quantity.

$\ MC(q_{1},q_{2})=\frac{TC(q_{2})-TC(q_{1})}{q_{2}-q_{1}}$

Marginal cost can either be thought of as the cost of producing the last unit of output or the cost of producing the next unit of output. Because of this, it's sometimes helpful to think of marginal cost as the cost associated with going from one quantity of output to another, as shown by q1 and q2 in the equation below.

To get a true reading on marginal cost, q2 should be just one unit larger than q1.

That said, as we consider smaller and smaller changes in quantity, marginal cost converges to the derivative of total cost with respect to quantity.

$\ MC(q_{1},q_{2})=\frac{dTC}{dQ}$

What if we need to calculate marginal cost as we go from one output point to a much bigger one?

For example, if the TC of producing 3 units of output is \$15 and the TC of producing 9 units of output is \$21, the marginal cost, simply put, is \$1.

Once we are considering a change in TC values due to a variation of quantity produced higher than 1 unity, are we applying the concept of average rate of change to measure the MC value?

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    $\begingroup$ "What if we need to calculate marginal cost as we go from one output point to a much bigger one?" You mean you want to estimate $MC(q_1)$ or $MC(q_2)$? There is no marginal cost for intervals. $\endgroup$ – Giskard Dec 29 '15 at 8:27
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I think you're getting things backwards. Marginal cost (at a certain quantity $q_0$) really only makes sense as the derivative of the total cost function with respect to quantity at point $q_0$: $$ MC(q_0)=\frac{dTC(q_0)}{dq}. $$ The interpretation of marginal cost as "the cost of producing the last unit of output or the cost of producing the next unit of output" is based on the (linear) approximation $$ \widehat{MC}^+(q_0)\approx\frac{TC(q_0+1)-TC(q_0)}{(q_0+1)-q_0}\quad\text{or}\quad \widehat{MC}^-(q_0)\approx\frac{TC(q_0)-TC(q_0-1)}{q_0-(q_0-1)} $$ This interpretation is usually introduced in undergraduate textbooks because of its intuitive appeal, not for its mathematical rigor. Imagine trying to explain the concept of marginal cost as the cost associated with an infinitesimal amount of change in quantity to a freshman who probably has no idea what "infinitesimal" means, let alone the concept of derivative.

A first problem related to interpreting marginal cost as the cost associated with the previous or the next unit of output is that producing the two units (the previous and the next) may imply different costs. This is illustrated by the formulas $\widehat{MC}^+$ and $\widehat{MC}^-$. Suppose total cost is quadratic, i.e. $TC(q)=q^2$. Then it can be easily verified that $\widehat{MC}^+(q_0)\ne \widehat{MC}^-(q_0)$, for any admissible $q_0$. This is problematic because we want the marginal cost (at a particular $q_0$) to be unique.

Second, and this relates to your question at the end, the concept of "unit" is somewhat arbitrary. And this arbitrariness makes measures such as $\widehat{MC}^+$ and $\widehat{MC}^-$ unstable as well. To illustrate, continue with the quadratic total cost example. When the unit of quantity is kg, the "marginal cost" at $1$kg--as you would refer to the cost associated with producing one extra unit--is $$\widehat{MC}^+(1)=2^2-1^2=3.$$ But what if we suddenly want to change the unit of quantity to gram (g)? Then the "marginal cost" at $1$kg, or $1000$g, would change as well: $$ \widehat{MC}^+(1)=(1.001)^2-1^2=0.002001. $$ To avoid such instability caused by changes in units of measurement, marginal cost is defined as a derivative, not as a difference.

If you want to measure the change in total cost due to change in some arbitrary quantity, simply apply the first formula in your question. But keep in mind that this cost is not marginal cost as an economist would understand it.

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  • $\begingroup$ In "Principles of Economics" by F. R. and B. B., marginal cost is believed to correspond to the average point of output over a interval [a,b] when the output produced changes from a to b. For example, if the TC of producing 10 units of output is \$25 and the TC of producing 20 units of output is \$45, MC is \$2. According to the book, \$2 is the MC that corresponds to the level of output of 15 units (because 15 lies between 10 and 20). Marginal Cost happens to be extrapolated in order to measure the “change in total cost due to change in some arbitrary quantity” Why is that so? $\endgroup$ – user6545 Dec 29 '15 at 14:53
  • $\begingroup$ I would think of Marginal Cost (when I want to measure the change in total cost due to changes in quantity from a to b) as being the Average Rate of Change. Then, I would apply the Mean Value Theorem to prove that there was at least one number c in the interval (a,b) (that is a < c < b) such that: $f'(c)=\frac{f(b)-f(a)}{b-a}$ Am I able to do such an assumption? $\endgroup$ – user6545 Dec 29 '15 at 17:04
  • $\begingroup$ @user6545: As denesp comments at your question, there is no marginal cost for intervals. To be clear, marginal cost is a function of quantity; that is, each quantity is associated with a (possibly different) marginal cost (level). The way you (or the textbook you use) are putting it suggests that each marginal cost is associated with a pair of quantities. This is why I said you seemed to be getting things backwards. $\endgroup$ – Herr K. Dec 29 '15 at 18:34
  • $\begingroup$ It's good that you mentioned the mean value thm. To calculate the marginal cost at quantity $c$ (i.e. $f'(c)$), we can look for an interval $(a,b)$ that contains $c$ and calculate the change in total cost (i.e. $f(b)-f(a)$) due to change in quantity in this interval (i.e. $b-a$). The MVT suggests that we can always find such an interval. This is presumably the rationale underlying the numerical example you gave in your first comment. But notice the use of language: I didn't say the marginal cost of $a$ and $b$, or the marginal cost from $a$ to $b$; I said the marginal cost at $c$. $\endgroup$ – Herr K. Dec 29 '15 at 18:36
  • $\begingroup$ @user6545: At best, you could say that, for any arbitrary interval $[a,b]$, the formula $\frac{f(b)-f(a)}{b-a}$ measures the marginal cost at some quantity $c\in(a,b)$. But this is much less precise. $\endgroup$ – Herr K. Dec 29 '15 at 18:36

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