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

<p>First the <strong>series approach</strong>.  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 $0 \leq q &lt; 1$. 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:</p>

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

<p>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.</p>

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

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

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

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

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

<p>Now the <strong>simpler way</strong>. 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:</p>

<p>$$\text{Variable cost = No. of units x Average marginal cost}$$</p>

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

<p>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.</p>

CC BY-SA 3.0
2017-11-20T22:12:59.657Z
2017-11-20T22:59:17.910Z
2017-11-20T22:59:17.910Z
Adam Bailey
<p>VC is not exactly the sum of MC but the <em>integral</em> of it. A sum is the discrete analog of an integral. I think you've been told this to help with intuitive comprehension. </p>

<p>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}</p>

CC BY-SA 3.0
2017-11-19T19:12:44.283Z
2017-11-21T15:41:22.063Z
2017-11-21T15:41:22.063Z
Herr K.
<p>We learned in class that variable cost is the sum of marginal cost. I am given the following question.</p>

<ol>
<li>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.</li>
</ol>

<p>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:</p>

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

<p>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</p>

<microeconomics>
Variable cost as the sum of marginal cost
user15248
# Variable cost as the sum of marginal cost

(Asked by: username_0 on 2017-11-19)

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

**Tags:** microeconomics

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## 2 answers

### Answer by username_1 on 2017-11-20. Score: 5

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.


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### Answer by username_2 on 2017-11-19. Score: 4

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}


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