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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?

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  • $\begingroup$ May be more suitable for Math Educators. $\endgroup$ – Simply Beautiful Art Oct 13 '17 at 22:25
  • $\begingroup$ The difference operator is just the discrete form of the derivative. Which one you work with generally depends on analytic tractability. For a linear cost they coincide. For a polynomial they are only close (ex. for $c(x)=x^2$ the derivative at $x=500$ is $c'(500)=2\times 500=1000$ while the difference operator gives $c(501)-c(500)=(500+1)^2-500^2=2\times 500+1=1001$). $\endgroup$ – lulu Oct 13 '17 at 22:30
  • $\begingroup$ Note: your students might be under the impression that the difference operator is more exact, which would be true if the cost function were exact. That, however, is generally not the case in practice. Usually the functions themselves are just approximations of reality, so the extra precision you get is unreliable. $\endgroup$ – lulu Oct 13 '17 at 22:31
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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).

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  • $\begingroup$ I'm not talking about $C$, I'm talking about its derivative. You approximate a function by going along the line with the derivative as its slope. See the diagram in the Wikipedia article. $\endgroup$ – Martin Argerami Oct 13 '17 at 22:38
  • $\begingroup$ There's no canonical answer to that. Derivatives are local, they don't really care about what happens far from them. The caveat is that to actually calculate the derivative you need to know the values of the function at infinitely many points, arbitrarily close to the point where you want to find the derivative. In the cost example, you only have discrete values, so you are likely to use a "discrete derivative". But in any case a derivative doesn't "see" what happens far from it. $\endgroup$ – Martin Argerami Oct 13 '17 at 22:52
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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: $$C'(100)\cdot 1+C'(101)\cdot 1 \approx \Delta C=C(102)-C(100).$$

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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.

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