Let x represent the big buses and y represent small buses. The large buses can carry 30 students and the small buses can carry 15 students. The total number of students are 450, this can be represented by the inequality:

30x + 15y ≤ 450

They are only 20 drivers, therefore only 20 buses can be used. It is represented by:

x + y ≤ 20

They are only 19 small buses and 18 large buses:

x ≤ 18

y ≤ 19

After plotting the graph, the minimum solution to the graph are at:

A (15,0), B(18,0), C(10, 10), D(18, 2).

The cost function is given as:

The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.

F(x, y) = 225x + 100y

At point A:

F(x, y) = 225(15) + 100(0) = $3375

At point B:

F(x, y) = 225(18) + 100(0) = $4050

At point C:

F(x, y) = 225(10) + 100(10) = $3250

At point D:

F(x, y) = 225(18) + 100(2) = $4250

The minimum cost is at point C(10, 10) which is $3250

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3 years ago