Also a tired car battery will have a higher impedance.
Maximum power transfer becomes a factor, if the voltage and internal resistance reach a point so that when the battery voltage is halved (by the current being drawn) the output power is under that required to start the engine, it won't work, whatever you do to it.
Your battery has an impedance of 6.6667mΩ.
Maximum power transfer will occur when the load impedance is equal to the battery's impedance, at this point the battery voltage will drop to half the open circuit voltage.
At what open circuit voltage will the battery be no longer capable of supplying the load with 3kW?
Calculate the voltage across a 6.6667mΩ load dissipating 3kW.
V = √(P×R) = √(3000×6.6667×10^-3) = 4.472V
This is the voltage across the load dissipating 3kW, the open circuit voltage will be twice this value so multiply by 2.
4.472×2 = 8.944V