Answer: minimum cost is $3250
You'll need 10 small buses and 10 large buses
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
thank you
