4
$\begingroup$

Why does the marginal cost equation (as the derivative of total cost equation) make predictions of variable costs that are very different from costs calculated using the Total Cost equation?

Marginal cost is simply the change in cost divided by the change in quantity.

MC = ΔC / ΔQ

However, marginal cost also can be computed using the derivative of the Total Cost function.

Suppose you have a short-term Total Cost equation for a production case in which no capital is used; labor is the only input.

TC = w * L

The production function is

Q = L^(1/3)   ... therefore
L = Q^3

And given that the w = 1, then

TC = Q^3

Therefore the Marginal Cost equation, as the derivative of the Total Cost equation, would be...

MC = 3Q^2

Of course, when Q = 0 then the TC equation and MC equation = 0. But raise Q to 1, and TC is now 1. Therefore, MC = ΔC / ΔQ = 1.

But the MC equation gives MC = 3.

The difference grows as Q increases more. When Q = 2, the TC equation returns 8, a cost change of 7, while the MC equation returns 12.

I understand mathematically why these are different, but I don't understand why both are supposedly correct and useful in economics.

Marginal analysis says to only produce a quantity if the marginal cost is less than or equal to the price at that quantity. But in this example, if the price were \$2, someone using the TC equation would produce the first unit for a profit of \$1 while someone using the MC equation would not produce the first unit because the MC equation predicts a \$1 loss. It seems like at least one of these methods lacks practical value. So, could someone explain the difference?

Improve this question
$\endgroup$
4
  • $\begingroup$ Seems to me you do not precisely know what marginal in marginal cost stands for. I recommend reading this: (It is not about production but consumption, but it gives an explanation of what marginal means.) en.wikipedia.org/wiki/Marginal_utility#Marginality $\endgroup$
    – Giskard
    Oct 9, 2015 at 19:44
  • $\begingroup$ "The cost of adding one more unit." That's what I thought marginal cost was. But I don't see why the MC equation gives me that when adding one more unit costs a different amount than what the MC equation returns. $\endgroup$
    – Bryan Gentry
    Oct 10, 2015 at 2:51
  • $\begingroup$ Marginal is about infinitesimally small units, marginal cost is a more involved mathematical concept, again I recommend reading the linked page. $\endgroup$
    – Giskard
    Oct 10, 2015 at 5:03
  • $\begingroup$ @BryanGentry See my question to Alecos answer below. Is that what you meant to ask? $\endgroup$
    – Quinn Culver
    Oct 10, 2015 at 13:50

1 Answer 1

Reset to default
2
$\begingroup$

For $Q_1 = 1, Q_2 =2$, you calculate

$$TC(Q_2) - TC(Q_1)=7$$ which is the additional (incremental) cost to produce the second unit of output, and then you compare this to

$$\frac {dTC}{dQ} |_{Q=Q_2} = 12$$

which is marginal cost when you are already at production level $Q_2$. From this observation alone, the two are not comparable.

As for practical value: assume marginal revenue is a straight line at $P=8$. What to do? The "marginal cost equals marginal revenue" leads to $3Q^2 \leq 8 \implies Q^*= \sqrt {8/3} \approx 1.633$. What's the problem with that?

Well, the product we are examining may not be divisible so much as to make incremental cost adequately close to marginal cost. This not-sufficient divisibility may force us to resort to integer-value calculus, and use the concept of incremental cost instead : produce the highest quantity for which incremental cost is lower or equal to incremental revenue - and we will obtain the correct answer $Q^* =2$.

Note: "sufficient divisibility" is not a strictly "physical constraint" but is judged also on economic considerations. If we are contemplating producing, say, cars to the range of hundreds of thousands, then, even if each car costs say 20 thousand euros, the pains and computational inconvenience of integer-valued calculus may not worth the small reduction in profits if (by using marginal calculus) we end up producing one unit more than optimal given indivisibility.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This helps somewhat, but it's still not clear to me how economists define MC as the cost of adding one additional unit. When MC = 12 at production level Q2, what does this mean? Adding one additional unit, Q3, brings total cost to 27. How does MC=12 relate to the cost incurred by adding this additional unit? $\endgroup$
    – Bryan Gentry
    Oct 10, 2015 at 2:56
  • $\begingroup$ And perhaps I should clarify my question some more. I am wondering why economics concerns itself so much with figuring out what the marginal cost is in practical applications. Why not just get TC(Q) and TC(Q+1) to see how much variable costs would increase if production is increased by 1 and then make a decision based on that? $\endgroup$
    – Bryan Gentry
    Oct 10, 2015 at 10:11
  • $\begingroup$ Shouldn't the question really be why is $TC(Q_2)-TC(Q_1) \neq \frac{dTC}{dQ}\vert_{Q=Q_1}$, since both are meant to represent the additional cost of producing an extra unit when already producing one unit? I could ask this as a separate questions since I want to know the answer. $\endgroup$
    – Quinn Culver
    Oct 10, 2015 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.