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 $TVC3(Q)$?


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 $Q > 50$.

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  • $\begingroup$ 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 $\endgroup$ – lala_12 Jul 6 '15 at 21:47
  • $\begingroup$ I am not sure what you denote by $TVC1$ and I don't know where 100 came from. $\endgroup$ – Giskard Jul 7 '15 at 4:43
  • $\begingroup$ I have edited my question - I forgot to add information about the last interval. $\endgroup$ – lala_12 Jul 7 '15 at 10:47
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    $\begingroup$ 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. $\endgroup$ – Giskard Jul 7 '15 at 11:46

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