<p>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}</p>

CC BY-SA 3.0
2015-04-06T11:09:05.897Z
Great thanks, sometime I complicate things for myself!
blzox
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2015-04-06T11:10:11.233Z
What would the 1st derivative of that be? (i.e MC)
blzox
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2015-04-06T11:11:23.587Z
Easy, it would just be MC = (r/a)*Q^((1-a)/a)
tadejsv
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2015-04-06T11:17:25.730Z
perfect thanks, now how would you interpret that in an economic context? Thanks a mill for your help!
blzox
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2015-04-06T11:28:39.817Z
@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.
FooBar
CC BY-SA 3.0
2015-04-06T14:20:02.650Z
Haha yeah I agree, looks too much like these are questions (b) and (c) from the same problem. The last question is too philosophical anyways, good luck with it!
tadejsv
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2015-04-06T11:02:49.370Z
2015-04-06T11:02:49.370Z
tadejsv
<p>In this question, only K is included and L is excluded how would I go about deriving it?<img src="https://i.sstatic.net/vRyiB.png" alt="enter image description here"></p>

<p>Total cost= Fixed costs + Average costs.</p>

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

<p>Thus $C(Q)= c_0 + rK^\alpha$</p>

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

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2015-04-04T17:28:03.120Z
Please show your work thus far or this is likely to be closed.
BKay
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2015-04-04T19:31:13.607Z
Last time I checked, univariate problems were easier to solve than multivariate...
Alecos Papadopoulos
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2015-04-06T10:02:05.250Z
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!
blzox
<mathematical-economics><cobb-douglas>
Derive a cost production function give prod f, only K
blzox
# Derive a cost production function give prod f, only K

(Asked by: username_0 on 2015-04-04)

In this question, only K is included and L is excluded how would I go about deriving it?![enter image description here](https://i.sstatic.net/vRyiB.png)

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.

**Tags:** mathematical-economics, cobb-douglas

> **Comments:**
> **username_1**: Please show your work thus far or this is likely to be closed.  
> **username_2**: Last time I checked, univariate problems were easier to solve than multivariate...  
> **username_0**: 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!  

---

## 1 answers

### Answer by username_3 on 2015-04-06. Score: 2

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}

> **Comments:**
> **username_0**: Great thanks, sometime I complicate things for myself!  
> **username_0**: What would the 1st derivative of that be? (i.e MC)  
> **username_3**: Easy, it would just be MC = (r/a)*Q^((1-a)/a)  
> **username_0**: perfect thanks, now how would you interpret that in an economic context? Thanks a mill for your help!  
> **username_4**: @username_0 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.  
> **username_3**: Haha yeah I agree, looks too much like these are questions (b) and (c) from the same problem. The last question is too philosophical anyways, good luck with it!  

---

