Originally posted by davegage
Yes, you are onto the right idea.
If you want to get the answer you would get if you considered infinitesimally small increments, don't limit yourself to whole mile increments, ie, on the first leg go 1000/19 = 52.63157..... miles, etc.
It shouldn't take too long before you arrive at the optimal answer...
EDIT: The idea behind this approach i ...[text shortened]... direction of Damascus without carrying his maximum load of 1000 KG. Hence it makes it optimal.
Ok. Well the best I could do was this (I don't have the patience to use anything
smaller than whole numbers):
Make 10 trips of 53 miles, which leaves you with 8993 kg at mile 53.
Make 9 trips of 59 miles, which leaves you with 7990 kg at mile 112.
Make 8 trips of 67 miles, which leaves you with 6985 kg at mile 179.
Make 7 trips of 76 miles, which leaves you with 5997 kg at mile 255.
Make 6 trips of 91 miles, which leaves you with 4996 kg at mile 346.
Make 5 trips of 111 miles, which leaves you with 3997 kg at mile 456.
Make 4 trips of 143 miles, which leaves you with 2996 kg at mile 599.
Make 3 trips of 200 miles, which leaves you with 1996 kg at mile 799.
Make 2 trips of the remaining 201 miles, which leave you with 1393 kg at Damascus.
I'm going to take a stab and guess that doing fractional milage increments saves
you an additional 50 kg at most (the difference if I landed on even 1000-kg increments
above).
Nemesio