Are these the constraints

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?What is the maximize income?
Are the constraints, x is for cars and y is for the buses

x+y<60 x<6 y<30, are these right?

x + y < or equal to 60..................total vehicles possible.

6x + 30y < or equal to 600............total space available.

or x + 5y < or equal to 100..........in simplified form.

We assume there must be some buses and cars

i.e. x > 0 and y > 0

Then find where the lines x + y = 60 and x + 5y = 100 intersect.

=> x + 5(60 - x) = 100

i.e. x + 300 - 5x = 100

=> 4x = 200

i.e. x = 50 => y = 10

Now, income is 2.5x + 7.5y

=> 2.5(50) + 7.5(10) = 125 + 75 = $200

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x+y<=60
no of buses=10 and number of cars=50 for maximize income
Maximum income=$7.5*10+$2.5*50=$75+$125=$200