Maximizing power from batteries to a DC Motor?
You want to run the motor at 38.2V * 2.32A = 89W. However you are limited to batteries weighing less than 150g that can be bought at a grocery store, which cannot deliver anywhere near that kind of power. Therefore the only choice you have is to slowly charge up some device (not a battery!) that can then be discharged at 89W for 10-20 seconds. 

The obvious choice for the storage device is a supercap. To deliver 89W for 20 seconds it needs to store at least 89*20 = 1780 joules of energy. In practice you will need more to compensate for conversion losses and to maintain a reasonable voltage throughout the discharge. Energy stored = 1/2CV2, so 2F at 50V (or 8F at 25V etc.) should be enough. Supercaps are usually rated at much lower voltages, but you can wire several in series/parallel to get the desired operating voltage and capacitance. 

How many caps? Using commonly available 10F 2.5V supercaps you need 20 in series to get 50V, but as total capacitance is divided by the number in series that only gives you 0.5F at 50V. So you will need 4 of these banks in parallel to get 2F - a total of 80 capacitors! 

Using a lower voltage won't help because the amount of energy stored is the same no matter how you connect the capacitors. However running a buck/boost regulator at a low input/output differential is generally more efficient than running a boost regulator from a much lower voltage. Using higher voltage also reduces current so you can use thinner wires. You need a regulator (buck, boost, or buck/boost depending on the operating voltage range of your capacitor bank) because voltage will drop as the capacitors discharge. 

To charge the capacitor bank you could use another switching regulator with current limiting. Your lithium batteries can deliver about 24W (6V @1A x4) so recharging the capacitor bank will probably take at least 1.5 minutes.