The area of a parking lot at an amusement park is 1310 square meters. a car requires 5 square meters and a bus requires 32 square meters of space. there can be at most 135 vehicles parked at one time. if the cost to park a car is $5.00 and a bus is $15.00, how many busses should be in the lot to maximize income? a. 24 c. 15 b. 111 d. 23

23 buses

Step-by-step explanation:

Let number of cars be x

Let number of buses be y

Now we are given that  A car requires 5 square meters

So, area required by x cars = 5 x sq.m.

A bus requires 32 square meters of space

So, area required by y buses = 32 y sq.m.

The area of a parking lot at an amusement park is 1310 square meters.

So, equation becomes:  ---1

There can be at most 135 vehicles parked at one time.

So, equation becomes:  --2

The cost to park a car is $5.00

Cost to park x cars = 5x

Cost to park a bus is $15

So, cost to park y buses = 15 y

So, Cost function =

Conditions:

Plot the equation 1 and 2 on the graph.So, the boundary points of the feasible region are (111.481, 23.519) , (262,0) and (135,0)

Now substitute the Points in the cost function to obtain the maximum income

Cost function =

At (111.481, 23.519)

Income :

At (262,0)

Income :

At (135,0)

Income :

So, when there are 111 cars and 23 buses in the parking lot then there will be maximum income .

Hence there are 23 buses in the lot to maximize income.

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