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September 20, 2014

September 20, 2014

Posted by **Anonymous** on Tuesday, August 31, 2010 at 1:25am.

I'm given the choices 10, 15, or 28

- PreCalc -
**drwls**, Tuesday, August 31, 2010 at 3:53amYou 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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