Question

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?

Answer 1

Answer:

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, $63x+7y \geq 252$

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

So, $2x+y \leq 22$

Rent of 1 bus = $1200

Rent of 1 van = $120

So, Cost = 1200x+120y

Objective function :  1200x+120y

Plot the lines on graph

$63x+7y \geq 252$  ---- Blue region

$2x+y \leq 22$ ----- 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