Variable cost as the sum of marginal cost

Question

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:

$$VC=(10+1/4)+(10+2/4)+...+(10+200/4)=2000+(200)(201)/8=7025$$

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

Answer 1

You can obtain variable cost as the sum of a series in this case because the given formula for marginal cost is linear. But there is a much simpler way.

First the series approach. The first term of your series was $10+1/4$ which is the marginal cost at $q=1$. This ignores the fact that, according to the formula, marginal cost is less than that whenever $0 \leq q < 1$. The first term should instead be the average marginal cost over the range $[0,1]$. Since marginal cost at $q=0$ is $10$, this is given by:

$$(1/2)[10 + (10 + 1/4)] = 10 + 1/8$$

Now the idea of a marginal cost for a quantity of less than one unit may seem nonsense if one thinks of marginal cost as the additional cost attributable to the marginal unit. But it does make sense for a good that can be produced in quantities that are not exact numbers of units of measurement (eg liquids, powders, measured in units of capacity or volume). In such a case, and given linearity, marginal cost can be defined as $\Delta C/\Delta q$, where the $\Delta$'s indicate changes in cost $C$ and $q$ respectively, and $\Delta q$ may be less than one unit.

By a similar logic, the second term in the series should be the average marginal cost over the range $[1,2]$ which is:

$$(1/2)[(10 + 1/4) + (10 + 2/4)] = 10 + 3/8$$

Continuing in this manner up to the final term which is the average marginal cost over the range $[199,200]$ we have:

$$VC = (10 + 1/8) + (10 + 3/8) \dots + (10 + 399/8) = 2000 + (100)(400)/8 = 7000$$

(The formula $(100)(400)$ comes from summing $100$ pairs: $1+399$, $3 + 397$, and so on.)

Now the simpler way. Given linearity, there is no need to consider intervals of one unit. The averaging formula can just be applied to the whole range $[0,200]$ as below:

$$\text{Variable cost = No. of units x Average marginal cost}$$

$$VC = (200)(1/2)[10 + (10 + 200/4)] = (200)(35) = 7000$$

Note however that neither of these methods would work if the formula for marginal cost were non-linear. Given a non-linear formula you would need to calculate an integral as illustrated in Herr K's answer.

Answer 2

VC is not exactly the sum of MC but the integral of it. A sum is the discrete analog of an integral. I think you've been told this to help with intuitive comprehension.

In general, \begin{equation} TC(Q)=FC+VC(Q) \end{equation} where $FC$ is independent of $Q$. To get MC, we take the derivative of TC with respect to $Q$: \begin{equation} MC(Q)=\frac{\mathrm dTC(Q)}{\mathrm dQ}=\frac{\mathrm dVC(Q)}{\mathrm dQ}. \end{equation} The second equality obtains because differentiating a constant always give you zero. Now to get VC from MC, we need the reverse operation of differentiation, which is integration, i.e. \begin{equation} VC(Q)=\int_0^{Q}MC(q)\,\mathrm dq. \end{equation} In your example, given $MC(Q)=10+\frac14Q$ and $Q=200$, \begin{equation} VC(200)=\int_0^{200}\left(10+\frac14q\right)\mathrm dq=\left.10q+\frac18q^2\right\vert_0^{200}=10(200)+\frac{200^2}{8}=7000 \end{equation}