# How do I calculate quantity to minimize long-run average total cost?

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?

• 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 May 26 '15 at 5:55
• 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 May 26 '15 at 7:44
• The total cost function should have a negative sign in front of 40, so the first derivative would be -40 +2Q = 0, and Q* = 20. – user6119 Oct 30 '15 at 18:10
• 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 Oct 30 '15 at 20:10

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

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

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