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Problem statement:

It costs: $$c(x)=x^{3}-6x^{2}+15x$$ dollars to produce x toys when 8 to 30 toys are produced and that $$r(x)=x^{3}-3x^{2}+12x$$ gives the dollar revenue from selling $x$ toys. Your toy shop currently produces $10$ toys a day. About how much extra will it cost to produce one more toy a day, and what is your estimated increase in revenue for selling 11 toys?

Solution:

The cost of producing one more toy a day when 10 are produced is about $c'(x)=3x^{2}-12x+15$ and $c'(10)=195$. The additional cost will be about $195$ dollars. The marginal revenue is: $r'(x)=3x^{2}-6x+12$. The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell $10$ toys a day, you can expect your revenue to increase by about $r'(10)=252$ dollars if you increase sales to 11 toys a day.

Why do we use calculus here?

I can already calculate the cost and the revenue of selling $11$ toys by calculating $c(11)$ and $r(11)$, and then I could calculate the increase in the cost and revenue by simply calculating $c(11)-c(10)$ and $r(11)-r(10)$. However, if I do that, the result is not the same.

I'm confused here. I'd be glad if you could help me with this one.

Thank you so much in advance!!! Cheers

tryst with freedom
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charlietan84
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    The calculus result is an approximation. You're right, you can directly and exactly calculate the required results, but they're trying to prove a point as some things need the first order derivative based approximation approach. – Deepak Jul 09 '21 at 10:23
  • Edited the question. Hopefully my edit still keeps the essence of what you wanted to ask. Else, feel free to roll back the question. – tryst with freedom Jul 09 '21 at 11:41
  • I mean the real answer is that economics uses calculus when a reasonable claim can be made that the goods under consideration are 'infinitely divisible' (i.e. quantities of sand) not discrete widgets. In the event one deals with discrete goods, there are formal tools for this but they require more math rather than less so tend to not be taught until a more advanced level. – Pete Caradonna Jul 09 '21 at 11:48
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    When you produce and sell hundreds or dozen thousands toys, and the costs and prices are not integer dollars but rather minor fractions (if a cost of running some machine is 50 dollars and it divides into ten thousands, then a unit cost of that machine 1/2 cent per toy); and you consider increase by 3,5% instead of one thousand fifty eight toys; then you certainly do not work in integers any more. Additionally all those values are estimates themselves, so a simple analytic approximation is much easier than exact arithmetics, even though a bit less precise. – CiaPan Jul 09 '21 at 11:50
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    What about inflation—is that not a ubiquitous concept in economics that calculus was designed to model? Quoting from Wikipedia: "When campaigning for a second term in office, U.S. President Richard Nixon announced that the rate of increase of inflation was decreasing, which has been noted as the first time a sitting president used the third derivative to advance his case for reelection." – Joe Jul 09 '21 at 11:54
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    A mold for the LEGO toys cost somewhere around 100,000 thousand dollars while the company produces tens of millions of pieces by this mold. So, if you divide $100,000/100,000,000 = 10^{-4}$ per piece. If you are working in these scales, really passing from discrete to continuous only make sense (at the end, the whole world is discrete as it is composed of atoms and true mathematical continuity does not exist). – William M. Jul 09 '21 at 16:46

3 Answers3

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It's a bad example.

In a course on economics you would likely make calculations resembling these, but it would more likely be for a factory producing ten thousand widgets a month than a shop producing ten toys a day. The marginal revenue at a certain level of production would be of interest. However, you would also be interested in questions such as how many toys to produce in order to maximize revenue. This occurs when the marginal revenue is zero.

If you make the reasonable assumption that your factory produces only an integer number of toys per time period, you can use discrete methods. But this would not seem like such an obvious assumption for a refinery producing fuel oil to be delivered by a pipeline. Either way, the methods of calculus give you adequate tools to find marginal rates and to find maxima or minima. If the maximum profit happens to be at a production rate of $20321.5$ toys per month, you can guess that the ideal rate of production is either $20321$ or $20322$ toys. Or make $20321$ toys one month and $20322$ the next. We're usually not interested in that exact an answer in that kind of economics.

What makes this a particularly bad example, however, is that the formulas are not particularly good economic models. In order to earn $r(x)$ dollars in total, you have to sell the toys for an average price of $\frac{r(x)}{x}$ dollars each. In this example, for $8 \leq x \leq 30,$ $\frac{r(x)}{x}$ is a strictly increasing function. That is, the more toys you make, the more you can sell each one for. Usually we're more interested in what happens when you sell so many toys that you saturate the market and the price starts dropping. (In fact it's not unusual to assume that the price is a strictly decreasing function of how many units you produce.) In that sense, this example is quite unrepresentative of the kind of problem it ought to be presenting.

David K
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The calculus method is vastly less computational effort than the method you describe.

You calculate $\frac{\mathrm{d}}{\mathrm{d}x} \left( r(x) - c(x) \right)$ once. This difference is a quadratic polynomial, so the computation is multiplying one constant by $2$ and writing down the resulting answer -- nearly effortless. Then evaluating this to find the cost of one additional unit of production requires one multiplication by a constant and one addition by a constant. This can, with a little practice, be done (or at least estimated) in your head.

Your method requires cubing both $10$ and $11$, squaring both $10$ and $11$, eight constant multiplication and eight additions/subtractions. This is not the sort of computation (or even estimate) nearly anyone can do in one's head. (When I competed in mental mathematics, this would have been at the edge of my abilities and would have required several restarts and re-computes to verify I hadn't mangled any of the many intermediate results.)

Equally useful... Time doing it both ways for $+1$, $+2$, and $+4$, that is, compare production of $10$ items with the options to produce $11$, $12$, and $14$ items. Now compare the time spent to the magnitude of the errors. Certainly, the "multiply the first derivative by the number of additional items" approximation method gets worse for large production changes and is low quality when the second derivative is large. (How bad ... is covered in Taylor series, with the keyword "error term", typically covered in "Calculus 2". Without diving into too many details, you can use the second derivative evaluated at the starting point and the change in production to bound how large an error you can make using this approximation method. For very many students, this is the first approximation method which comes with a computable (in-)accuracy bound, vastly improving on just waving hands and saying "approximately". In fact, if the bound is too large (so a large error is possible) you also gain the tools to compare the computational cost of using more derivatives in the approximation versus picking a starting point closer to the target production versus restarting the computation from scratch.)

Finally, the problem is unrealistic in that one seldom has an analytic expression for costs and revenue. One may have a point estimate for costs and revenue at the current production level and estimates of "elasticity" in these numbers. Again avoiding too many details, this corresponds to knowing the first derivatives of cost and revenue. So, in addition to letting you use an easy estimation technique and presenting an approximation technique for which you can compute bounds on the badness of the approximation, you also start practicing using the sort of data you will actually be able to obtain in real life.

Eric Towers
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You can avoid calculus here because the numbers are small and toys are discrete. If $x$ instead represented a continuous quantity (or one so large that a continuous approximation makes sense) then calculus would be needed.

user432944
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