Question

A bus company has contracted with a local high school to carry 450 students on a field trip.

Answer 1

Using 11 of the large buses and 8 of the small buses and the total would be $3275

Answer 2

Answer: minimum cost is $3250

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

$\begin{cases}0 \le x \le 18\\0 \le y \le 19\\x+y \le 20\\30x+15y \le 450\\\end{cases}$

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