The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport students, requires chaperones, and costs $ to rent. Each van can transport students, requires 1 chaperone, and costs $ to rent. Since there are students in the senior class that may be eligible to go on the trip, the officers must plan to accommodate at least students. Since only parents have volunteered to serve as chaperones, the officers must plan to use at most chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation costs? What are the minimal transportation costs?

We require 2 buses and 18 vans for minimal cost

The minimal transportation costs $4560

Step-by-step explanation:

Let x represents the no. of buses and y represents the no. of vans

We are given that  Each bus can transport 63 students, requires 2 chaperones, and costs $1200 to rent. Each van can transport 7 students, requires 1 chaperone, and costs $120 to rent.

We are also given that the officers must plan to accommodate at least 252 students.

So,

We are also given that the officers must plan to use at most 22 chaperones.

So,

Rent of 1 bus = $1200

Rent of 1 van = $120

So, Cost = 1200x+120y

Objective function :  1200x+120y

Plot the lines on graph

 ---- Blue region

----- Green region

So, The points of the feasible region are (4,0),(2,18) and (11,0)

At (4,0)

Transportation cost : 1200x+120y=1200(4)+120(0)=4800

At (2,18)

Transportation cost : 1200x+120y=1200(2)+120(18)=4560

At (11,0)

Transportation cost : 1200x+120y=1200(11)+120(0)=13200

So, The cost is minimum at (2,18)

So, We require 2 buses and 18 vans for minimal cost

Hence  the minimal transportation costs $4560

answer: finding the extract measurements and knowing the points

step-by-step explanation: