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. At most 80 vehicles can park at one time. If the cost to park a car is $2.00 and the cost to park a bus is $6.00, how many buses should be in the lot to maximize income?

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

You 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.

15 buses

To maximize income, we need to find the number of buses that should be in the parking lot.

Let's assume the number of buses in the parking lot is "b".

From the given information, we know that the area of the parking lot is 805 square meters, a car requires 5 square meters, and a bus requires 32 square meters.

We can set up the equation:
5c + 32b ≤ 805
where "c" is the number of cars and "b" is the number of buses.

Next, we know that at most 80 vehicles can park at one time, so we have the constraint:
c + b ≤ 80

To maximize income, we need to find the maximum value for 6b, since each bus costs $6.00 to park.

Now, let's solve the problem step-by-step:

1. Rearrange the inequality to isolate "c":
c ≤ 805 - 32b / 5

2. Substitute this inequality into the constraint equation:
805 - 32b / 5 + b ≤ 80

3. Simplify the inequality:
805 - 32b + 5b ≤ 400

4. Combine like terms:
-27b ≤ -405

5. Divide by -27, remembering to reverse the inequality:
b ≥ -405 / -27
b ≥ 15

Since the number of buses cannot be negative, we can conclude that "b" should be at least 15 to maximize income.

Therefore, the correct answer is 15 buses.

To maximize the income, we need to determine the maximum number of buses that can be parked in the lot considering the given constraints.

First, let's find out how many cars can be parked in the lot. We have the total area of the parking lot, which is 805 square meters, and each car requires 5 square meters of space. So, the number of cars that can be parked at one time is:

Number of cars = Total area of the parking lot / Area required per car
Number of cars = 805 square meters / 5 square meters
Number of cars = 161 cars

Now, let's calculate how many square meters of the parking lot are occupied by these 161 cars:

Total area occupied by cars = Number of cars * Area required per car
Total area occupied by cars = 161 cars * 5 square meters
Total area occupied by cars = 805 square meters

To find out the remaining area in the parking lot for the buses, subtract the area occupied by cars from the total area of the parking lot:

Remaining area for buses = Total area of the parking lot - Total area occupied by cars
Remaining area for buses = 805 square meters - 805 square meters
Remaining area for buses = 0 square meters

Since we have no remaining area for the buses, it means that we cannot park any buses in the parking lot while also accommodating the maximum number of cars.

Therefore, the correct answer is 0 buses.

None of the given choices (10, 15, or 28) are valid for maximizing income in this scenario.