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In this question, only K is included and L is excluded how would I go about deriving it?enter image description here

Total cost= Fixed costs + Average costs.

Since the variable input costs r per unit, the variable costs is r times the number of units rQ, hence $VC= rK^\alpha$.

Thus $C(Q)= c_0 + rK^\alpha$

Is the above solution correct? IS there a more mathematical way of doing it, one that involves grpahs etc.

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  • $\begingroup$ Please show your work thus far or this is likely to be closed. $\endgroup$ – BKay Apr 4 '15 at 17:28
  • $\begingroup$ Last time I checked, univariate problems were easier to solve than multivariate... $\endgroup$ – Alecos Papadopoulos Apr 4 '15 at 19:31
  • $\begingroup$ If both K and L were present, it would have been much easier to derive it, however with L not being part of the production function, it gets a bit tricky! $\endgroup$ – blzox Apr 6 '15 at 10:02
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Here you have to express $K$ in terms of $Q$, since the cost depends on the number of units of capital employed and not on the number of products produced. You have \begin{equation} Q = K^a \Leftrightarrow K = Q^{\frac{1}{a}} \end{equation} so your cost function will be \begin{equation} C(Q) = c_0 + rK(Q) = c_0 + rQ^{\frac{1}{a}} \end{equation}

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  • $\begingroup$ Great thanks, sometime I complicate things for myself! $\endgroup$ – blzox Apr 6 '15 at 11:09
  • $\begingroup$ What would the 1st derivative of that be? (i.e MC) $\endgroup$ – blzox Apr 6 '15 at 11:10
  • $\begingroup$ Easy, it would just be MC = (r/a)*Q^((1-a)/a) $\endgroup$ – tadejsv Apr 6 '15 at 11:11
  • $\begingroup$ perfect thanks, now how would you interpret that in an economic context? Thanks a mill for your help! $\endgroup$ – blzox Apr 6 '15 at 11:17
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    $\begingroup$ @blzox it seems like you're expecting him to solve the whole problem set? Be cautious with the number and type of followup-questions you pose in comments to someone who already answered your initial question. $\endgroup$ – FooBar Apr 6 '15 at 11:28

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