Misha and Dontavious were asked to solve the following problem using linear programming.
A local school district can operate a large bus for $300 per day and a small bus for $200 per day. The district needs to arrange transportation for at least 360 students for a field trip and has enough chaperones to have one chaperone per bus for up to 7 buses. Each large bus can hold 60 students and each small bus can hold 45 students. What is the minimum cost for the transportation for the field trip?

Misha decided the following represented this situation:

Let x = number of large buses

Let y = number of small buses

Objective Fx: C=300x+45y

Constraints:

200+60y≥360

x+y≤7

x≥0

y≥0

Dontavious decided the following represented this situation:

Let x = number of large buses

Let y = number of small buses

Objective Fx: C=300x+200y

Constraints:

60x+45y≥360

x+y≤7

x≥0

y≥0

Decide which student set up the correct Objective function and Constraints and using that information answer the following:

a) What are the vertices for the correct feasible region?

b) What is the minimum cost for transportation for the field trip and how did you arrive at this cost?

Dontavious wrote the appropriate inequalities. cost is $300 for large buses and $200 for small buses. and the constraints are the number of students per bus.

a) The feasible region is where the solution spaces overlap: this creates a small triangular area with the vertices at (6, 0), (7, 0) and (3, 4).

(6, 0), (7, 0), (3, 4)

b) 3 large buses and 4 small buses would cost 3*300 + 4*200 = $1700

6 large buses and 0 small buses would cost 6*300 = $1800 dollars

The cheapest solution would be 3 large buses and 4 small buses which would cost $1700