The area of a 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 180 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?: (

To maximize the income should be 30 buses and 150 cars

Step-by-step explanation:

Let

x-----> the number of cars

y ----> the number of bus

we know that

------> inequality A

----> inequality B

The function of the cost to maximize is equal to

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),(150,30),(180,0)

Verify

(0,53) --->

(150,30) --->

therefore

To maximize the income should be 30 buses and 150 cars

Givens

To solve this problem we need to create a table to order all this information and express it as a system of inequations.

                     Car           Bus     Total Capcity

Sq. Meters      5c           32b            1710

N° vehicles      c              b                180                

Therefore, the inequalities are

The expresion which represents the income is

, because a car is $2.00 and a bus is $6.00.

Now, we first need to find the critical points of the solution of the inequality system, which is attached. Observe that the only points that can be a solution is (150,30), because there can't be just car or just buses.

Then, we replace this point in the income expression

Therefore, in order to maximize incomes, we need to park 150 cars and 30 buses, to make $480 income.

answer: i cant see the picture

step-by-step explanation:

can u take a better picture

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation: