This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach":

The total cost $C$ of producing $x$ units of some item is a function of $x$. (Of course physically $x$ takes on only positive integral values, but it is convenient to think of it as taking on all real values in some domain.) Economists use the term marginal cost for the rate of change of $C$ with respect to $x$. Suppose that $C = 5x^2 + 15x + 200$, what is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?

Since the question asks for the rate of change at a specific value of $x$, we compute $C' = 10x + 15$. Then the marginal cost when $x = 15$ is equal to $165$.

I am still trying to understand how this would be the cost of the 16th unit though. Is it that since the rate of change of $C$ at $x = 15$ is 165, moving from 15 units to 16 units adds that rate to the overall cost? Similar to how if an object is moving at 165 mph, then in one hour it would travel 165 additional miles. Can this even be compared to an object in motion though? The object could easily change speeds within the hour, but there can be no fractional units produced.

Using the method of increments on the total cost formula yields \begin{align} C + \Delta C &= 5(x +\Delta x)^2 + 15(x + \Delta x) + 200 \\ &= 5x^2 + 10x \Delta x + 5(\Delta x)^2 + 15x + 15 \Delta x + 200 \\ \end{align}

Then \begin{align} \Delta C = C + \Delta C - C&= 10x \Delta x + 5(\Delta x )^2 + 15 \Delta x \\ \end{align}

Setting $x = 15$ and $ \Delta x = 1$, we have \begin{align} \Delta C = 10* 15 * 1 + 5 * 1^2 + 15* 1 = 170 \end{align}

Why is this value of $\Delta C$ not the cost of the 16th unit?


What you're calling the method of increments is the slope of the chord from 15 to 16, $(C(15+\Delta)-C(15))/\Delta $ with $\Delta=1$, and the derivative is what happens as $\Delta \rightarrow 0$, giving you the slope of the tangent line to the graph at 15. They're not going to be the same unless the function is constant.

When calculus gets used in economics it is usually just a story. It's not meant to be taken literally. Most results have "non-calculus" versions that come from linear programming or lattice arguments, but the arguments are more tedious and abstract, despite making more practical sense.

  • $\begingroup$ The text seems to state with certainty that the cost of the 16th item is 165. Why is the cost of the 16th item equal to the derivative? $\endgroup$ – Iyeeke Jun 26 '20 at 15:32
  • $\begingroup$ Imagine you are running 20 miles. In between mile 15 and 16, you burn 170 calories in total, but the moment you start mile 15, you are burning 165 calories for an infinitesimal period of time. It is the same thing here. Imagine you are pulling a beer tap and filling up a pitcher. To get from oz 15 to 16, it costs 170 dollars. But at the start of the 15th oz, how much does it cost? 165 dollars times an infinitesimal drop of beer. $\endgroup$ – user762914 Jun 26 '20 at 23:31
  • $\begingroup$ Would it then be reasonable to say that the cost of the 16th oz of beer is 165, or the calories burned during the 16th mile is equal to 165? That is how the text seems to be interpreting this value. $\endgroup$ – Iyeeke Jun 27 '20 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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