Question

The area of a
parking lot is 600 square meters. A car requires 6 square
meters. A bus requires 30 square meters. The attendant can handle only
60 vehicles. If a car is charged $2.50 and a bus $7.50, how many of each
should be accepted to maximize income?

Answer 1

The number of cars should be 100 and that of bus be 0.

x is the number of cars and y is the number of buses.

Area: 6x + 30y <= 600 simplifies to x + 5y <= 100

Total money collected: 2.50x + 7.50 y …..this is what you want to maximize

OR you want to maximize 2.50 (x+ 3 y)

OR you want to maximize (x+3y)

Replacing the value of x from the Area equation in the above price, we get

100 - 5y + 3y is the maximum amount you can collect

So you want to maximize 100 - 2y. This happens when y is 0.

When y is 0; x will be 100 and you will collect 250.

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