Q: Given ATC=3Q^{2}-2Q+4/Q, find FC and VC.

A: There are several different ways to solve this problem, since we know
several different relations among the cost curves:

TC=VC+FC, FC=TC(when Q=0), ATC=TC/Q, AVC=VC/Q, AFC=FC/Q, ATC=AFC+AVC, and
MC=dTC/dQ=dVC/dQ.

The easiest way to proceed is to first calculate TC and then use the the fact
that FC=TC(when Q=0) and finally solve for VC=TC-FC

To find TC: ATC=TC/Q so TC=ATC*Q. **TC **= (3Q^{2}-2Q+4/Q)*Q
= **3Q ^{3}-2Q^{2}+4**

To find FC: FC=TC(when Q=0). **FC **= 3(0)^{3}-2(0)^{2}+4
= 0-0+4 = **4**

To find VC: VC=TC-FC. So **VC** = 3Q^{3}-2Q^{2}+4-4
=** 3Q ^{3}-2Q^{2}**