The student activities department of a community college plans to rent buses and vans for a spring-break trip. Each bus has 40 r
egular seats and 1 handicapped seat; each van has 8 regular seats and 3 handicapped seats. The rental cost is $350 for each van and $975 for each bus. If 320 regular and 36 handicapped seats are required for the trip, how many vehicles of each type should be rented to minimize cost
2 answers:
Answer:
6 buses and 10 vans
Step-by-step explanation:
It is convenient to solve this graphically. The problem setup is ...
minimize 975b + 350v subject to ...
40b +8v ≥ 320 . . . . . regular seats required
1b +3v ≥ 36 . . . . . . . .handicapped seats required
where b and v are the numbers of buses and vans, respectively.
The constraint lines intersect at (b, v) = (6, 10). 6 buses and 10 vans will provide the required number of seats at minimum cost.
_____
At any other point in the feasible region, the cost line (green) will be farther from the origin, hence the cost will be greater.
In the attached, x represents buses; y represents vans.
Answer:
10 vans and 6 buses.
Step-by-step explanation:
Minimize cost = 350*x + 975*y
subject to
40*y + 8*x >= 320 (regular seats needed)
y + 3*x >= 36 (handicap seats needed)
x >= 0
y >= 0
where
x, number of vans rented
y, number of buses rented
This is a linear programing problem. In this kind of problem, it is known that the optimal solution is on a vertix of the feasible region. In the figure attached, feasible region can be seen. For the vertices costs are
vertex cost
(0, 36) $35100
(10, 6) $9350
(40, 0) $14000
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