Could you help me with this? The question is too long to fit here.

Question

The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 49 students, requires 2 chaperones, and costs ​$1,400 to rent. Each van can transport 7 students, requires 1 chaperone, and costs $110 to rent. Since there are 392 students in the senior class that may be eligible to go on the​trip, the officers must plan to accommodate at least 392 students. Since only 26 parents have volunteered to serve as​chaperones, the officers must plan to use at most 26 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation​ costs? What are the minimal transportation​ costs? Thanks for the help!

Answer 1

Let we rent x buses and y vans.

Need to accommodate at least 392 students: 49x + 7y ≥ 392

Can use at most 26 chaperones: 2x + y ≤ 26

Also, there are restrictions: x ≥ 0, y ≥ 0

Want to minimize transportation costs: 1400x + 110y → min

This is a linear programming problem, but it can be solved graphically.

Graph the inequalities: 49x + 7y ≥ 392

2x + y ≤ 26

x ≥ 0, y ≥ 0

Intersection of all half-planes is the triangle with vertices:

(8, 0) (8 buses), (13, 0) (13 buses), and (6, 14) (6 buses, 14 vans).

Optimal solution must be in one of these vertices, so we can calculate transportation costs for each vertex and compare them.

For (8, 0) the cost is 1400×8 = $11,200

For (13, 0) the cost is 1400×13 = $18,200

For (6, 14) the cost is 1400×6 + 110×14 = $9,940

So, the officers should rent 6 buses and 14 vans. The minimal transportation cost is $9,940.