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Apples

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kamil9876:
Part 2 can be done by extending that optimal strategy:

i know i can get 1000 apples to B with 4000 at A by splitting it up into two lots of 2000. However if there exists a better method, it must be one where the apples at A are between 4000 and 3000 since 3000 gives a max that is less than 1000. Hence we can write the number of apples at A as 3000+a.

that means initially there are 4 apples being eaten per km if we drag across the 4 loads. This occurs for a distance until the a apples are eaten. Hence:



Now the next 1000 apples must be eaten such that we eat 3 apples per km until we get to position :

hence:


The final 1000 apples must be eaten for a distance , 2 per km this time:

Solving these equations backwards to find a gives Hence apples at A.

Your version can be done by applying this too.

evaporade:
one more to go

kamil9876:
okay

3000=999*3+3 hence he has 4 loads at the start. loses 4apples/km ==>

3 apples per km:



2 apples per km:
2(x_3-x_2)=999
x_3=833.25

so far he has 999 apples left. and 166.75 km to go, hence apples delivered is: 832.25. so 832 delivered and 0.25 for his smoko.

evaporade:
you made a slip at the start, otherwise well done.

Now what about the camel and the bananas?

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