My students don't appreciate the marginal cost function, please help,

Question

Given a cost function c(x), and knowing the cost for some large enough x, say, 500 units, we can find the incremental cost of producing the 501st unit, by computing c'(500). And then we can also find the incremental cost of producing the 502nd unit by computing c'(501). So, with this information, we know the incremental costs for producing an extra 2 units, from 500 to 502.

My students weren't satisfied at all with this example, because many of them were perplexed about why I couldn't just compute c(502) - c(500) to get my answer. I was stumped and told them that I don't think they are wrong, but at the moment, I did not have a better answer for them to convince them that the marginal cost function + using derivatives is a good thing.

What can I tell them to highlight the usefulness of the marginal cost function and derivatives?

Answer 1

The point is that you might not know $C(502)$, and you want to estimate it; by using the derivative, you only need to know the function $C$ up to $501$ to estimate it at $502$. A company wants to know the cost of the item in advance of producing (it helps with all kind of decisions). If you are keeping track of what the cost of each item is, you can predict the cost of the next item (or any item, but precision decreases the further you go along).

Answer 2

A common application of marginal cost: Under suitable conditions on the revenue and cost functions, profit maximization occurs iff marginal cost equals marginal benefit.

Setting the marginal cost function equal to the marginal benefit function and solving for a quantity is easier than comparing the marginal cost and benefit of lots of possibilities.

Answer 3

The students should understand that the limit definition of the derivative (marginal cost) is: $$MC(x)=C'(x)=\lim_{\Delta x\to 0} \frac{\Delta C}{\Delta x}.$$ Note that it is defined for $\Delta x\to 0$ (instantenous change). However, if the limit is omitted, then the formula becomes an approximation: $$C'(x)\approx \frac{\Delta C}{\Delta x} \Rightarrow \Delta C\approx C'(x)\Delta x.$$ Note that the smaller the change of $x$, the more accurate the change of $C$.

Hence: $$C'(100)\cdot 1+C'(101)\cdot 1 \approx \Delta C=C(102)-C(100).$$