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First, I can find a cost function where fixed cost is involved.

  • Eight units cost $\$300$; fixed cost is $\$60$.

I got $C(x)= 30x + 60$

However, for a problem like

  • Twelve units cost $\$445$; $50$ units cost $\$1585$.

I don't think my solution process is right. I get the slope like always, which is $30$, and create the $C(x)$ from one of the points \begin{align*} p-445 & = 30(q-12)\\ p-445 & = 30q-360\\ p & = 30q + 85\\ C(x)& = 30x+85 \end{align*}

Now, the solution says the function I've created is wrong. Can anyone tell me what I'm doing wrong?

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  • $\begingroup$ Your solution is correct. There is an error in the solution key. $\endgroup$ – N. F. Taussig Jun 9 '17 at 10:41
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For the second,

Slope should be $\frac{1585-445}{50-12}=\frac{1140}{38}=\$30$ per unit.

Now point slope, $p-445=30(q-12)$

$p-445=30q-360$

$p=30q+85$

$C(x)=30x+85$

Your solution seems fine to me.

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    $\begingroup$ For the first one, we have a slope of $$\frac{300 - 60}{8} = \frac{240}{8} = 30$$ You forgot to take into account the fixed costs. $\endgroup$ – N. F. Taussig Jun 9 '17 at 10:33
  • $\begingroup$ Oops! @taussig haha $\endgroup$ – Saketh Malyala Jun 9 '17 at 15:51

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