Answer:
Explanation:
From the given information;
Let consider p to be the number of large buses and q to be the number of small buses.
The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students.
The linear inequality represent the students with respect to the total students s:
30p + 15q ≥ 450 -----(1)
They are only 20 drivers available on the day of the field trip, therefore only 20 buses can be used. So,
p + q ≤ 20 ----- (2)
Also , There are only 19 small buses and 18 large buses:
∴
0 ≤ p ≤ 18
0 ≤ q ≤ 19
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
In order to minimize the cost of transporting all 450 students, let z be the minimal cost. i.e
z = 225p + 100q
From equation 1 and 2 ; if we plot them graphically on the graph, the following points of intersection were obtained as shown in the sketch below, in which the shaded region lies the answer.
The point of intersection between equation (1) and (2) is (p,q) = (10,10)
From the critical point in the sketch of our graph attached below, the following values of z can be determined.
Point(s) Value for Z
(15,0) 15 × 225 = 3375
(18,0) 18 × 225 = 4050
(18,2) (18 × 225 )+ (2 × 100 ) = 4250
(10,10) (10 × 225) + (10 × 100) = 3250
Thus ; the minimum cost of transporting all 450 students = $3250