Algebra 2 Math Problem

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 only handle 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

let c = # of cars

let b = # of buses

Limitation on vehicles

c + b ≤ 60

Limitation of space

6c + 30b ≤ 600 - OR -

c + 5b ≤ 100

Subtract

c + 5b ≤ 100

c + b ≤ 60

----------------

4b ≤ 40

b ≤ 10

If there are 10 buses, then there can be at most 50 cars.

10(7.5) + 50(2.5) = $200 revenue

Answer 2

If we can maximize the amount of space used between cars and buses then that should give us the maximum income.

The amount of space has to be less than or equal to 600 sq meters. Since each car takes 6 sq meters and each bus take 30 square meters

6C + 30B ≤ 600

Also, since the attendant can only handle 60 vehicles

C + B ≤ 60

Now we have two inequalities and two variables. Since we're maximizing space and vehicles, we'll make these equations:

6C + 30B = 600

C + B = 60

We can use the second equation to find C in terms of B and then substitute that value in the first equation

C = 60 - B

6(60-B) + 30B = 600

Now you can solve for B and then solve for C