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

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

  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.

300pi

step-by-step explanation:

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p