Question

C=q^3+1.5q^2+9q......

Find AC .....

MC

AVC

AFC..?

Answer 1

$AC = \frac{TC}{Q}$

$AC = \frac{{q^{3}+1.5q^{2}+9q}}{q}$

$AC = q^{2}+1.5q+9$

$MC = \intop$ $\frac{dTC}{dQ}$

$MC = 3q^2 + 3q + 9$

$AVC = \frac{TVC}{Q}$

In this case; $AVC = AC$ because there is no fixed cost

$AVC=q 2 +1.5q+9$

$AFC = \frac{TFC}{Q}$

In this case; $TFC = 0; thus AFC = 0$