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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? Please help :(

Respuesta :

calculista

Answer:

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

[tex]5x+32y\leq1,710[/tex] ------> inequality A

[tex]x+y\leq 180[/tex] ----> inequality B

The function of the cost to maximize is equal to

[tex]C=2x+6y[/tex]

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) ---> [tex]C=2(0)+6(53)=\$318[/tex]

(150,30) ---> [tex]C=2(150)+6(30)=\$480[/tex]

therefore

To maximize the income should be 30 buses and 150 cars

Ver imagen calculista

Answer:

Givens

  • The area of the parking lot is 1710 square meters.
  • A car requires 5 square meters.
  • A bus requires 32 square meters.
  • There can be a maximum of 180 vehicles.
  • The cost for a car is $2.00.
  • The cost for a bus is $6.00.

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

[tex]5c+32b\leq 1710\\c+b\leq 180[/tex]

The expresion which represents the income is

[tex]I=2c+6b[/tex], 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

[tex]I=2c+6b\\I=2(150)+6(30)=300+180=480[/tex]

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

Ver imagen jajumonac

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