It is none of them.
If you have access to these two plants and you want to produce y units, then you would like to produce it in a way that minimizes cost of production:
min a^2 + 2b^2
subject to a + b = y
Solving the above optimization problem with respect to a and b, where a is the number of units produced in Plant A and b is the number of units produced in Plant B, we get
a = (2/3) y and b = (1/3) y
and joint cost is a^2 + 2b^2 = [(2/3) y]^2 + 2[(1/3) y]^2 = (6/9) y^2 = (2/3) y^2