Why does marginal cost (derivative of total cost) differ from variable cost at each level?

Question

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?

Answer 1

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.