Question

The area of parking lot is 1710 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. There can be at most 189 vehicles parked at one time. Of 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?

Answer 1

Answer:

To maximize the income should be 28 buses and 160 cars

Step-by-step explanation:

Let

x-----> the number of cars

y ----> the number of bus

we know that

$5x+32y\leq1,710$ ------> inequality A

$x+y\leq 189$ ----> inequality B

The function of the cost to maximize is equal to

$C=2x+6y$

Solve the system of inequalities by graphing

The solution is the shaded area

see the attached figure

The vertices of the solution are

(0,0),(0,53),(160,28),(189,0)

Verify

(0,53)

$C=2(0)+6(53)=\ 318$

(160,28)

$C=2(160)+6(28)=\ 488$

therefore

To maximize the income should be 28 buses and 160 cars

Answer 2

Answer:

There should be 30 buses in the lot to max out income

Step-by-step explanation: