Given the following formula for total cost:
$C=x^3-6x^2+18x$
Find the value of $x$ that minimizes the average cost, and show that at this value of $x$ the marginal cost and average cost are equal
$Solution:$
Given the formula for total cost $C=x^3-6x^2+18x$, to get the formula for average cost which we will denote by $\bar{C}$ you have to divide $C$ by $x$.
Therefore we have:
$\bar{C}=x^2-6x+18$
We have to minimize this function. So let's take the derivative and solve for its critical points:
$\bar{C}'=2x-6=0$
$\rightarrow x=3$
If we wanted to, we could confirm that this value of $x$ is where a minimal value occurs by doing the second derivative test OR by cutting up the number line and checking for the sign of $\bar{C}'$ on each of the regions and then drawing a conclusion from there.
However, it turns out that in section 3.4/3.5 these techniques are unnecessary, because when the questions in this section ask for a maximum or minimum value, and you do the obvious first step of finding the critical points for the first derivative, that the critical points you get are always going to give you the type of extrema you are looking for. Sometimes you will get two critical points, and if (for example) the question is asking for a max, just plug them both in and see which gives you a larger value.
ANYWAY,
The second part of this question is asking to show that at the value of $x$ that minimizes the average cost, the average cost and the marginal cost are equal to eachother. Formula for marginal cost is given by $C'$; so we need to show that $C'(3) = \bar{C}(3)$
$C'=3x^2-12x+18$
$\rightarrow C'(3)=9$
$\bar{C}= x^2-6x+18$
$\rightarrow \bar{C}(3)=9$
In fact, you can prove this result generally. Let $C(x)$ be a differentiable cost function with $C(0)=0$. Then the average cost function is $\displaystyle \bar{C}(x) = \frac{C(x) - C(0)}{x - 0} = \frac{C(x)}{x}$ and $\displaystyle [\bar{C}(x)]' = \frac{xC'(x) - C(x)}{x^2}$. If $x_m$ minimizes the average cost then it must be the zero of its derivative, that is $\bar{C}(x_m) = 0$. So
$\displaystyle \frac{x_mC'(x_m) - C(x_m)}{(x_m)^2} = 0 \Longrightarrow x_mC'(x_m) = C(x_m) \Longrightarrow \frac{C(x_m)}{x_m} = C'(x_m) \Longrightarrow \bar{C}(x_m) = C(x_m)$