Linear Programming By: Jordan Pickett and Diana Vlassenko

The officers of a highschool senior class are planning to rent buses and vans for a class trip. Each bus can transport 40 students, requires 3 chaperones, and costs $1200 to rent. Each van can transport 8 students, requires 1 chaperone, and costs $100 to rent. The officers must plan to accommodate at least 400 students. Since only 36 parents have volunteered to serve as chaperones, the officers must plan to use at most 36 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation costs? What are the minimal transportation costs? Transportation Scenario

Variables and Information Buses=X 40 students, 3 chaperones $1200 to rent 400 students 36 parents Vans=Y 8 students, 1 chaperone $100

Constraints 40x + 8y ≥ 400 3x + 1y ≤ 36 Objective function: We are trying to minimize so we can get the lowest transportation cost possible. So we would want to get the least amount of buses and vans that would make it the lowest price to rent. Using the previous constraints listed, and remembering x= buses and y= vans, the below objective function will get us that answer. 1200x + 100y= cost

Linear Programming Transportation Graph POINTS: (10,0) (12,0) (8,10)

Calculations POINTS: (10,0) (12,0) (8,10) OBJECTIVE FUNCTION: 1,200x+100y=cost SOLUTIONS: 1,200(10)+100(0)=12,000 1,200(12)+100(0)=14,000 1,200(8)+100(10)=10,600

ANSWER: 1,200(8)+100(10)=10,600 This would be the answer meaning we need 8 buses and 10 vans to seat the 400 students and 36 parents using the least amount of money possible (minimizing). The other solutions were: 1,200(10)+100(0)=12,000 1,200(12)+100(0)=14,000 Those solutions would be more expensive, so they wouldn’t be the answer we’re looking for considering we want the cheapest one. We got the solutions by taking the coordinate points that surrounded the feasible region and plugging them into the objective function. Then we took the smallest cost number and made that our answer.

In Conclusion… In order to minimize transportation costs, the high school would need to rent 8 buses and 10 vans, which would equal $10,600.