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. Th
ere 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 :(
2 answers:
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
------> 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)
<em>Verify</em>
(0,53) --->
(150,30) --->
therefore
To maximize the income should be 30 buses and 150 cars
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
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.
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