A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental cost for a large bus is $800 and $600 for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost. Let x = the small buses and y= the big buses.

5 small buses and 4 large buses are needed to get a total cost of $6200.

Let x represent the number of small buses and y represent the number of big buses.

Since there are 400 students and 10 buses of 50 seats each and 8 buses of 40 seats, hence:

40x + 50y = 400   (1)

Also there are 9 drivers available, hence:

x + y = 9        (2)

Therefore x = 5, y = 4

Then cost for a large bus is $800 and $600 for the small bus, hence:

Total cost = 600x + 800y = 600(5) + 800(4) = $6200

Therefore 5 small buses and 4 large buses are needed to get a total cost of $6200.

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Step-by-step explanation:

Let us assume the total number of small buses needed = x

The capacity of 1 small bus  = 40

So, the capacity of x buses  = 40(x)  = 40 x

Let us assume the total number of big buses needed = y

The capacity of 1 big bus  = 50

So, the capacity of y buses  = 50(y)  = 50 y

Also, the total students travelling = 400

So, the number of students traveling by (Small bus + Big bus)  = 400

⇒ 40 x + 50 y = 400 (1)

Also, the total number of drivers available  = 9

⇒ x +  y = 9   (2)

Also, x  ≤ 8,   y ≤ 10

Now, solving both equations, we get:

40 x + 50 y = 400 (1)

x +  y = 9  ⇒ y = (9-x) put in (1)

40 x + 50 y = 400  ⇒  40 x  + 50 (9-x)  = 400

or, 40 x  + 450 - 50 x  = 400

or, - 10 x  =- 50

or, x  = 5 ⇒ y = (9-x)  = 9- 5 = 4

Hence the number of small buses used = 5

The number of big buses used  = 4

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