Question

The student activities department of a community college plans to rent buses and vans for a spring-break trip. Each bus has 40 regular 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

Answer 1

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 2

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