Posted by Anonymous on Tuesday, August 31, 2010 at 1:25am.
You could set this up as a caldculaus problem and find the maximum-income number of buses, or just calculate and compare the income for each the three choices given.
With 10 buses, you have 10x32 = 320 m^2 for buses and 805-320 = 485 m^2 for 485/5 = 97 cars. Total income is 60 + 194 = $254 . However, the number of vehicles exceeds tha allowed .
With 15 buses, you have 480 m^2 for buses and 325 m^2 for 65 cars. The income is 90+130 = $220
Now consider the case of 28 buses. Chances are the total income wil be less than the 15 bus case.
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