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Apples

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dcc:
evaporade, could you post the steps that one must go through to arrive at the destination with 834? I can't seem to replicate your answer :|

evaporade:
First dropoff point at 1000/3 km from A, next dropoff point at 500 km from the first.

Have to make 3 forward trips (1000 apples each trip) from A to the first dropoff. Total distance = 1000/3 x 3 = 1000 km. .: 1000 apples eaten and 2000 apples left at the first dropoff point.

Have to make 2 forward trips (1000 apples each trip) from the first dropoff to the next dropoff. Total distance = 500 x 2 = 1000 km. .: 1000 apples eaten and 1000 apples left at the second dropoff point.

The remaining distance = 1000 - 1000/3 - 500 = 166 2/3 km.

One final forward trip (1000 apples) of 166 2/3 km to B. .: 166 apples eaten, 834 delivered.

I let you think about why those two dropoff points were chosen.

kamil9876:
This is similair to shinny's solution. He takes 1000 apples and carries them 1km, then goes back for the next 1000 and carries them 1km, then the last lot carries 1km. So by travelling 1km he loses 3 apples. He loses 1000 apples by travelling 1000/3 km. At this point, he only requires two back trips. And so he just takes one lot, goes back for the other and carries that 1km as well, so he is losing 2 apples per km. until 1000/2=500 km later where he has 1000 apples, and now he only does not have to return, but just go straight there, hence the total apples he comes back with is 1000-(1000 - (1000/3 + 500)).

Note how this is equivalent to evaporade's method, just in a different order ie: instead of going back for each load after 1km, why not 0.5km? 0.1, or even the whole 1000/3 since it's all quite arbitrary. This is because you will lose those apples in the second load by one 333.3 km of trip as much as you will by 333.3 lots of those smaller 1km trips.

evaporade:
To see one really understand, try to explain

2.) What's the minimum value of apples that he should start of with (rather than 3000) at A so that he can deliver 1000 apples to B?

and

Another question: If the capacity of the truck is 999, what is the max delivery to B?

kamil9876:
Oh and as a proof:

imagine if he only had to travel 1km. The obvious solution would be 3 lots of 1000, rather than 6 lots of 500 of anything like that. of course 1 lot of 3000 would be best but we cannot have this, hence minimum number of possible trips ())should always be taken. This minimum is occurs until 1000/3.

In comparison to my attempt, or any other one for that matter, is that we went over the same domain too many times. e.g: i went over the (0,500) domain 3 times, whereas evaporade only went over the (333.3,500) domain twice and so he saved 166.7 apples apples there. Going minimum lots by 1km is a safe way to guarantee unnecesary trips, and by looking ahead we can see it has the same outcome as 3 big journeys to 1000/3.

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