| Solution: (1) The maintenance and repair fee for 100 days of use is: 100+101+102+...+199=14950 rounds, so the average daily consumption is: (14950+500000)÷1 00=5149.5 rounds. (2) Let the use of n days scrapped. Then **** need to pay maintenance and repair costs are: 100 + 101 + 102 + ... + (n + 99) = n2 + n (rounds).? Therefore, the average daily consumption is: (n2+ n+500000) ÷ n = n+ + . When n=, i.e. n=1000 days the average daily consumption is the least, so it is best to use 1000 days for scrapping. |
| Slightly |