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](https://tex.z-dn.net/?f=63x%2B7y%20%5Cgeq%20252)
We are also given that the officers must plan to use at most 22 chaperones.
So, ![2x+y \leq 22](https://tex.z-dn.net/?f=2x%2By%20%5Cleq%2022)
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