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.

  • $\begingroup$ Please show your work thus far or this is likely to be closed. $\endgroup$
    – BKay
    Apr 4, 2015 at 17:28
  • $\begingroup$ Last time I checked, univariate problems were easier to solve than multivariate... $\endgroup$ Apr 4, 2015 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, 2015 at 10:02

1 Answer 1


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}

  • $\begingroup$ Great thanks, sometime I complicate things for myself! $\endgroup$
    – blzox
    Apr 6, 2015 at 11:09
  • $\begingroup$ What would the 1st derivative of that be? (i.e MC) $\endgroup$
    – blzox
    Apr 6, 2015 at 11:10
  • $\begingroup$ Easy, it would just be MC = (r/a)*Q^((1-a)/a) $\endgroup$
    – tadejsv
    Apr 6, 2015 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, 2015 at 11:17
  • 1
    $\begingroup$ @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, 2015 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.