(2) Assume that it will be scrapped after n days. Then the maintenance fee to be paid by * * * is:100+1+102+...+(n+99) = N2+n (rounded). So the average daily consumption is: (n2+ n+500000)÷n= n++.
When n=, that is, n= 1000 days, the average daily consumption is the least, so it is best to use 1000 days for scrapping.
leave out