<p>Your are right. You have to minimize the average cost.</p>

<p>$$c(Q)=\frac{C(Q)}{Q}=6000 +40Q+Q^2$$</p>

<p>Calculate the first derivative and set it equal to zero:</p>

<p>$ c'(Q)=40+2Q=0 $</p>

<p>Solve this equation for $Q$. Denote the optimal value as $Q^*$. $Q^*$ can be a local maximum or a local minimum</p>

<p>If $c''(Q^*)&gt;0$, then you have found the local minimum.</p>

<p>The local minimum is also the absolute minimum, because</p>

<p>$$\lim_{Q \to \infty } 6000 +40Q+Q^2=\infty$$</p>

<p>$$\lim_{Q \to -\infty } 6000 +40Q+Q^2=\infty$$</p>

CC BY-SA 3.0
2015-05-26T05:55:34.103Z
2015-05-26T07:42:23.530Z
2015-05-26T07:42:23.530Z
callculus42
<p>I have a formula for the long-run total cost curve,</p>

<p>$$TC(Q) = 6000Q + 40Q^2 + Q^3$$</p>

<p>and I'm trying to find the quantity that minimizes the long-run average total cost. </p>

<p>I assume I'm trying to find the value $Q$ (quantity) that produces the total cost ($TC(Q)$) where $\frac{TC(Q)}{Q}$ is the lowest. But how do I do this short of trial and error?</p>

CC BY-SA 3.0
2015-10-30T20:10:39.473Z
The question as it is does not result in a positive value for $Q^*$. The author might have mis-wrote the question, but we can't say that there *should* have been a negative sign in front of the 40.
Kitsune Cavalry
CC BY-SA 3.0
2015-05-26T05:55:34.840Z
The question itself is fine. I feel like I might be misreading it though. Since all the values are positive, shouldn't the lowest average cost occur when Q=0?
Jamzy
CC BY-SA 3.0
2015-05-26T07:44:49.017Z
This function does not have a minimum for strictly positive values of $Q$, and this is Economics and so "negative quantity" makes no sense -even zero quantity is uninteresting. My suspicion is that at least one of the terms should appear with a minus sign before it.
Alecos Papadopoulos
<microeconomics><long-run><cost>
How do I calculate quantity to minimize long-run average total cost?
Doug Smith
# How do I calculate quantity to minimize long-run average total cost?

(Asked by: username_0 on 2015-05-26)

I have a formula for the long-run total cost curve,

$$TC(Q) = 6000Q + 40Q^2 + Q^3$$

and I'm trying to find the quantity that minimizes the long-run average total cost.

I assume I'm trying to find the value $Q$ (quantity) that produces the total cost ($TC(Q)$) where $\frac{TC(Q)}{Q}$ is the lowest. But how do I do this short of trial and error?

**Tags:** microeconomics, long-run, cost

> **Comments:**
> **username_2**: The question itself is fine. I feel like I might be misreading it though. Since all the values are positive, shouldn't the lowest average cost occur when Q=0?  
> **username_3**: This function does not have a minimum for strictly positive values of $Q$, and this is Economics and so "negative quantity" makes no sense -even zero quantity is uninteresting. My suspicion is that at least one of the terms should appear with a minus sign before it.  
> **username_1**: The question as it is does not result in a positive value for $Q^*$. The author might have mis-wrote the question, but we can't say that there *should* have been a negative sign in front of the 40.  

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## 1 answers

### Answer by username_4 on 2015-05-26. Score: 6 | Accepted answer

Your are right. You have to minimize the average cost.

$$c(Q)=\frac{C(Q)}{Q}=6000 +40Q+Q^2$$

Calculate the first derivative and set it equal to zero:

$ c'(Q)=40+2Q=0 $

Solve this equation for $Q$. Denote the optimal value as $Q^\*$. $Q^\*$ can be a local maximum or a local minimum

If $c''(Q^\*)>0$, then you have found the local minimum.

The local minimum is also the absolute minimum, because

$$\lim\_{Q \to \infty } 6000 +40Q+Q^2=\infty$$

$$\lim\_{Q \to -\infty } 6000 +40Q+Q^2=\infty$$


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