Question

If total cost function is given by $C=a+bx+c{x}^2$, where x is the quantity of output show that
$\frac{d}{dx}\left(AC\right)=\frac{1}{x}(MC-AC)$, where MC is the marginal cost and AC is the average cost.

Answer 1

Here, $C=a+bx+c{x}^2$
Quantity $=x$
$\therefore AC=\frac{C}{x}=\frac{a+bx+c{x}^2}{x}=\frac{a}{x}+b+cx$
And $MC=\frac{d}{dx}\left(C\right)=\frac{d}{dx}(a+bx+c{x}^2)=b+2cx$.
$\therefore \frac{1}{x}(MC-AC)=\frac{1}{x}(b+2cx-\frac{a}{x}-b-cx)$
$=\frac{1}{x}(cx-\frac{a}{x})$
$\therefore \frac{1}{x}(MC-AC)=c-\frac{a}{x^2}$ ........... (i)
$\frac{d}{dx}\left(AC\right)=\frac{d}{dx}(\frac{a}{x}+b+cx)$
$\frac{d}{dx}\left(AC\right)=-\frac{a}{x^2}+c$ ........... (ii)
From (i) and (ii), we get
$\frac{d}{dx}\left(AC\right)=\frac{1}{x}(MC-AC)$.