The area of a parking lot is 805 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. There can be at most 80 vehicles parked at one time. If the cost to park a car is $2.00 and a bus is $6.00, how many buses should be in the lot to maximize income?

B = number of buses parked

C = number of cars parked

We can write 3 formulas:

Formula 1: Total number of vehicles cannot exceed 80.

B+C<=80

The maximum income will be when it is 80

Formula 2: The space occupied cannot exceed 805

5C+32B<=805

The maximum income will be when it is 805

So let’s try to get 805 m^2 with 80 vehicles at the same time

B+C=80 - > C=80-B

5C+32B=805 - > C = (805-32B) / 5

Let’s do the equalization method:

80-B = (805-32B) / 5

5 * (80-B) = 805-32B

400-5B=805-32B

32B-5B=805-400

27B=405

B=405/27=15

C=80-15=65

Let’s verify:

5*65+32*15=805 - > OK

Formula 3: Income=6*15+2*65=220

If we put more buses, it doesn’t fit. Let’s try with 1 more bus and 1 less car:

5*34+32*16=832>805 - > NOT OK

If we put less buses, it’s less income. Let’s try with 1 less bus and 1 more car:

Income=6*14+2*66=216 - > Not so convenient.

So the optimum is 15 buses and 65 cars.