Q: f(L,K)=3L+2K. In the short run Labor is fixed at L=2. If w=2 and v=1 what is the cheapest way to make 12 units of output in the short run?

A: Since L=2 is fixed in the short-run, to make 12 units of output, it
must be that f(2,K)=3(2)+2K=12. Or that 6+2K=12. This means that
2K=6 or K=3. Therefore the optimal bundle is L=2 and K=3. The cost
of this is wL+vK=2(2)+1(3)=7.

*You should note that the cheapest way to produce 12 units in the long-run is
with 6 units of K and 0 units of L which would only cost 6. *