We learned in class that variable cost is the sum of marginal cost. I am given the following question.

  1. Assume you are a profit-maximizing firm in a perfectly competitive market. Marginal cost is MC(qi) = 10 + qi/4 and the market price for you output is P= $60.

Calculate variable costs. So the teacher goes ahead and solves for the area of a trapezoid to get a variable cost of 7000 (when q=200) and that's fine. But why do I get a different answer when I try to calculate the sum of a series? My attempt:


Clearly 7025 doesn't equal 7000, so what is going on here? Also I thought MC isn't supposed to include fixed costs so is the equation MC(qi)=10+qi/4 only defined when qi is at least 1? Thanks


TLDR: This is the difference between "discrete" and "continuous" variables.

Integration and differentiation are tools dealing with continuous variables. That is, variables that can take on any real value. In calculus, $x$ can be integers such as $1$ or $2$, but it can also be fractions such $1/2$ or $4/3$ or $2.0678358723$. It can even be irrational values such as $\sqrt 2$ or $\pi$. This is explicit in how these tools are defined, what it is they mean. If you have some target value you are interested in, say for example what happens to some function $f(x)$ when $x$ is near $2$, then to use calculus tools to investigate $f(x)$ at $2$, there can be no limitation on how close $x$ is allowed to get to $2$. $x$ may be $2.1$ or $2.01$ or $2.00000000000001$ or even closer.

However, there are many situations - particularly in economics, but also plenty of other cases - where variables cannot take on any value. When dealing with the number of units produced or sold, or with the number of people meeting some criterion, only integers values are sensible. In some cases, fractions are allowed, but only certain ones. A price function with values in dollars can sensibly take on values of $\$2.25$ and $\$3.07$, but not $\$3.007$. Only fractions out of $100$ are allowed.

Such variables are called "discrete". Unlike continuous variables, there is a limitation on how close two values can be. For counts, that limit is $1$. For dollars, that limit is $\$0.01$. Now to reiterate, the tools of calculus apply to continuous variables. They definitions do not apply to discrete variables. But economics - particularly business mathematics - almost always is about discrete variables. So what is the point of "Business Calculus"?

Calculus may not work directly with discrete variables, but it has a lot - a VERY LOT - to say about how functions behave, and that behavior can have important implications for economic models and running businesses. What is the best price to sell your product at, and why? What will be the result of pricing away from this optimal value? There are many question that calculus can answer.

But in order to do calculus, you need continuous variables. So to apply calculus, we have to turn the actual discrete variables into continuous ones. Fortunately, this is usually not hard to do - particularly for the sort of problems found in textbooks and classrooms, where you magically know what the function actually looks like. (There are good reasons for teaching like this I won't get into, but it isn't realistic.) For example, your marginal cost function above. $qi$ and $MC(qi)$ are discrete in their meaning, but there is no reason that the formula $$MC(qi) = 10 + \frac{qi}4$$ has to be restricted to discrete values. The formula still makes sense if one allows $qi$ and $MC$ to take on all values. Only the meaning of those numbers is lost away from the discrete values. But calculus can now be applied. However, with a caveat: because calculus results are affected by the meaningless extension to continuous variables, everything calculus tells you is only an approximation.

That is what is happening here. Your series is the actual "variable cost". The area under the trapezoid - the integral - is only an approximation of it. In the graph, when you summed the series, what you calculated was the area in yellow (extended to $qi = 200$), but the area your teacher calculated was only the region under the line. The $25$ difference is the total area above the line.

Area comparison of discrete versus continuous trapezoid

  • $\begingroup$ Thanks for the great answer! This makes a lot of sense. $\endgroup$ – Jonathan Nov 19 '17 at 23:03

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