<p>You lost a constant at
$$TVC(Q)=\int{MC(Q)}dQ$$
because for any $\alpha \in \mathbb{R}$ the derivative of $TVC(Q) + \alpha$  will be $MC(Q)$.</p>

<p>The way to get this constant: <br>
If there is no quasi-fixed cost then $TVC(0) = 0$. From this and by calculating $TVC(Q)$ for $Q \leq 50$ you will get the value for $TVC(50)$. (Seems to be 3750.) Because $TVC(Q)$ is a continuous function
$$
\lim_{Q \to 50^{+}} TVC(Q) = TVC(50)
$$
where by $Q \to 50^{+}$ I mean that $Q$ converges to 50 from above (you take the right hand side limit). This, together with $$TVC(Q)=\int{MC(Q)}dQ$$ should give you $TVC(Q)$ for $Q &gt; 50$.</p>

CC BY-SA 3.0
2015-07-06T21:47:30.160Z
Thank you! So to get the constant 1250, would it be correct to estimate TVC, for 50 < Q <=100, as TVC=int(MC(Q),Q) + (TVC1(100) - TVC1(50))?

where TVC1 is for Q <= 50
lala_12
CC BY-SA 3.0
2015-07-07T04:43:31.157Z
I am not sure what you denote by $TVC1$ and I don't know where 100 came from.
Giskard
CC BY-SA 3.0
2015-07-07T10:47:49.367Z
I have edited my question - I forgot to add information about the last interval.
lala_12
CC BY-SA 3.0
2015-07-07T11:46:13.857Z
I find this quite bad practice, because now the answer is out of sync with the question both in notation and in content. I will not edit my answer anymore, but if you understand what I wrote I am confident you can figure out the answer to your current question.
Giskard
CC BY-SA 3.0
2015-07-06T20:58:18.293Z
2015-07-06T20:58:18.293Z
Giskard
<p>I need to estimate the function expressions for $TVC$ and $AVC$ from the $MC$ function for different amounts of quantity $(Q)$</p>

<p>From the solutions I have, that</p>

<p>For $Q \leq 50$ </p>

<p>$MC1(Q) = -Q + 100$</p>

<p>$TVC1(Q) = -0.5 Q^2 + 100Q$ </p>

<p>$AVC1(Q) = -0.5Q+100$</p>

<p>$\\$</p>

<p>For $50 &lt; Q \leq 100$ </p>

<p>$MC2(Q) = 50$</p>

<p>$TVC2(Q) = 50Q + 1250$ </p>

<p>$AVC2(Q) =50 + \frac{1250}{Q}$</p>

<p>$$\\$$</p>

<p>For $Q &gt; 100$ </p>

<p>$MC3(Q) = 0.5Q$</p>

<p>$TVC3(Q) = 0.25 Q^2 + 3750$ </p>

<p>$AVC3(Q) = 0.25Q + \frac{3750}{Q}$</p>

<p>How can I estimate the constants in $TVC2(Q)$ and $TVC3(Q)$?</p>

<microeconomics><cost>
Estimate total variable cost function (TVC) and average variable cost function (AVP) from marginal cost function (MC)
lala_12
# Estimate total variable cost function (TVC) and average variable cost function (AVP) from marginal cost function (MC)

(Asked by: username_0 on 2015-07-06)

I need to estimate the function expressions for $TVC$ and $AVC$ from the $MC$ function for different amounts of quantity $(Q)$

From the solutions I have, that

For $Q \leq 50$

$MC1(Q) = -Q + 100$

$TVC1(Q) = -0.5 Q^2 + 100Q$

$AVC1(Q) = -0.5Q+100$

$\\$

For $50 < Q \leq 100$

$MC2(Q) = 50$

$TVC2(Q) = 50Q + 1250$

$AVC2(Q) =50 + \frac{1250}{Q}$

$$\\$$

For $Q > 100$

$MC3(Q) = 0.5Q$

$TVC3(Q) = 0.25 Q^2 + 3750$

$AVC3(Q) = 0.25Q + \frac{3750}{Q}$

How can I estimate the constants in $TVC2(Q)$ and $TVC3(Q)$?

**Tags:** microeconomics, cost

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## 1 answers

### Answer by username_1 on 2015-07-06. Score: 1

You lost a constant at
$$TVC(Q)=\int{MC(Q)}dQ$$
because for any $\alpha \in \mathbb{R}$ the derivative of $TVC(Q) + \alpha$ will be $MC(Q)$.

The way to get this constant:   
If there is no quasi-fixed cost then $TVC(0) = 0$. From this and by calculating $TVC(Q)$ for $Q \leq 50$ you will get the value for $TVC(50)$. (Seems to be 3750.) Because $TVC(Q)$ is a continuous function
$$
\lim\_{Q \to 50^{+}} TVC(Q) = TVC(50)
$$
where by $Q \to 50^{+}$ I mean that $Q$ converges to 50 from above (you take the right hand side limit). This, together with $$TVC(Q)=\int{MC(Q)}dQ$$ should give you $TVC(Q)$ for $Q > 50$.

> **Comments:**
> **username_0**: Thank you! So to get the constant 1250, would it be correct to estimate TVC, for 50 < Q <=100, as TVC=int(MC(Q),Q) + (TVC1(100) - TVC1(50))?  where TVC1 is for Q <= 50  
> **username_1**: I am not sure what you denote by $TVC1$ and I don't know where 100 came from.  
> **username_0**: I have edited my question - I forgot to add information about the last interval.  
> **username_1**: I find this quite bad practice, because now the answer is out of sync with the question both in notation and in content. I will not edit my answer anymore, but if you understand what I wrote I am confident you can figure out the answer to your current question.  

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