Question

A school district needs to arrange transportation for at least 360 students for a field trip. The district can use large buses or small buses. The large buses hold 60 students each and have an operating expense of $300 for the day. The small buses hold 45 students each and have an operating expense of $200 for the day. The district only has chaperones for, at most, seven buses. What is the minimum cost for the transportation for the field trip if each bus has chaperones?
$1,400
$1,600
$1,700
$1,800

Answer 1

To find the minimum cost of transportation for at least 360 students, we can set up a system based on the number of large buses (L) and small buses (S) used, along with the constraints given.

1. **Capacity Constraints**:
Each large bus holds 60 students, and each small bus holds 45 students. Thus, the total capacity can be modeled as:
\[
60L + 45S \geq 360
\]

2. **Bus Constraint**:
The total number of buses used must be 7 or fewer:
\[
L + S \leq 7
\]

3. **Cost Function**:
The cost associated with using the buses is given by the operating expenses:
\[
\text{Total Cost} = 300L + 200S
\]

We want to minimize the total cost under the given constraints.

### Analyzing the Feasible Solutions
Let's explore different combinations of large and small buses based on the constraints.

- **Case 1:** Use 0 large buses (L = 0)
\[
45S \geq 360 \implies S \geq 8 \quad \text{(not feasible, since } L + S \leq 7\text{)}
\]

- **Case 2:** Use 1 large bus (L = 1)
\[
60(1) + 45S \geq 360 \implies 45S \geq 300 \implies S \geq \frac{300}{45} = 6.67 \implies S \geq 7 \quad (L + S \leq 7 \text{ is satisfied for } S=6 \text{ only)}
\]
This is not feasible since \(1 + 7 = 8\) exceeds the bus limit.

- **Case 3:** Use 2 large buses (L = 2)
\[
60(2) + 45S \geq 360 \implies 120 + 45S \geq 360 \implies 45S \geq 240 \implies S \geq \frac{240}{45} = 5.33 \implies S \geq 6 \quad (L + S = 2 + 6 > 7\text{, not feasible)}
\]

- **Case 4:** Use 3 large buses (L = 3)
\[
60(3) + 45S \geq 360 \implies 180 + 45S \geq 360 \implies 45S \geq 180 \implies S \geq 4 \quad (L + S = 3 + 4 = 7 \text{ is feasible)}
\]
- Total Cost:
\[
300(3) + 200(4) = 900 + 800 = 1700
\]

- **Case 5:** Use 4 large buses (L = 4)
\[
60(4) + 45S \geq 360 \implies 240 + 45S \geq 360 \implies 45S \geq 120 \implies S \geq \frac{120}{45} = 2.67 \implies S \geq 3 \quad (L + S = 4 + 3 = 7 \text{ is feasible)}
\]
- Total Cost:
\[
300(4) + 200(3) = 1200 + 600 = 1800
\]

- **Case 6:** Use 5 large buses (L = 5)
\[
60(5) + 45S \geq 360 \implies 300 + 45S \geq 360 \implies 45S \geq 60 \implies S \geq 1.33 \implies S \geq 2 \quad (L + S = 5 + 2 > 7, \text{ not feasible)}
\]

### Summary of Costs
The feasible combinations we analyzed yielded these results:

- 3 large buses and 4 small buses: **Cost = $1,700**
- 4 large buses and 3 small buses: **Cost = $1,800**

The minimum cost is therefore:
\[
\boxed{1700}
\]