A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students.
There are only 20 drivers available on the day of the field trip.
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
Determine the minimum cost of transporting all 450 students.

You'll need 10 small buses and 10 large buses

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Explanation:

x = number of large buses

y = number of small buses

The restrictions for x and y are 0 <= x <= 18 since we have at most 18 large buses (and x can't be negative) and 0 <= y <= 19 for similar reasoning (this time we have up to 19 small buses to work with).

Furthermore, x+y <= 20 represents the idea that only 20 drivers are available. Each bus gets a driver and x+y is the total number of buses in operation. We want x+y to be 20 or smaller.

We'll also be dealing with the equation 30x+15y = 450

30x = number of students that go on the x large buses

15y = number of students that go on the y small buses

30x+15y = total number of students = 450

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The objective cost equation is

C = 225x+100y

with 225x being the cost of operating all the large buses and 100y being the cost of operating the small buses. We want to get C as small as possible.

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Graph this system of inequalities

Note that the corner points shown in the diagram below are

A = (0, 0)

B = (0, 19)

C = (1, 19)

D = (10, 10)

E = (15, 0)

Plug each of those (x,y) values into the cost function C = 225x+100y

Ignore (0,0), since this is a trivial solution. Operating no buses at all will yield a cost of 0 dollars, but this isn't what we're after. Either x or y must be positive.

You should find that (0, 19) leads to the min value of C as it produces the smallest output for C. However, notice that

1 small bus = 15 students

19 small buses = 19*15 = 285 students

15+285 = 300 which is less than 450

So we'll need a few large buses. Focus on points that are on the boundary line of 30x+15y <= 450. Those points are D and E. Point D's coordinates will lead to C = 3250 being the smallest cost value. This time we're able to transport all 450 students

10 small buses = 10*15 = 150 students

10 large buses = 10*30 = 300 students

150+300 = 450 total