These are electric cars.
Edit: The efficient approach would take symmetry into account as shown by @R. J. Mathar
These are electric cars.
Edit: The efficient approach would take symmetry into account as shown by @R. J. Mathar
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There is at least one simple symmetric solution (not necessarily the maximum, but optimization problems often have extrema at the symmetric choices): Let all 6 cars drive from the initial point (all range 1000) by 1000/6 so their remaining range is down to $5\times 1000/6$. Recharge 5 of the cars equally with the power of the 6th (the 6th is halted there) so we have 5 cars at distance 1000/6 from the origin with range 1000 left. After another distance 1000/5 recharge 4 of the cars with the power of the 5th and this leaves 4 cars at distance 1000/6+1000/5 with range 1000 left. Recursively ... the last car can reach $1000/6+1000/5+1000/4+1000/3+1000/2+1000 = 1000\times 49/20 = 2450$.