Question

The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 80 ​students, requires 4 ​chaperones, and costs ​$1 comma 100 to rent. Each van can transport 10 ​students, requires 1​ chaperone, and costs ​$120 to rent. Since there are 480 students in the senior class that may be eligible to go on the​ trip, the officers must plan to accommodate at least 480 students. Since only 40 parents have volunteered to serve as​ chaperones, the officers must plan to use at most 40 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation​ costs? What are the minimal transportation​ costs?

Answer 1

• 2 buses and 32 vans
• $6040 is the minimum cost

Explanation:

We can write inequalities related to the two constraints (at least 480 students, at most 40 chaperones), identify the feasible solution space, and locate the solution with the least cost. Since there are only two transportation types to choose from, we can solve the problem graphically. Or, we can reason through it.

Let x represent the number of buses in the mix, and y represent the number of vans. The constraint on capacity is ...

80x +10y ≥ 480 . . . . . . . we must transport at least 480 students

4x + y ≤ 40 . . . . . . . . . . . we must require at most 40 chaperones

The objective function we want to minimize is ...

c(x, y) = 1100x +120y

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The attached graph shows these inequalities plotted. They overlap in a wedge-shaped area with vertices at (2, 32), (6, 0), and (10, 0). Clearly, we can ignore the point (10, 0), as the feasible solution (6, 0) is obviously cheaper.

Evaluating the objective function at the possible solution points, we find ...

c(2, 32) = 1100·2 +120·32 = 6040

c(6, 0) = 1100·6 +120·0 = 6600

To minimize transportation costs, 2 buses and 32 vans should be rented. The minimum cost is $6040.

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Solution by reasoning

The cost per student for a bus is $1100/80 = $13.75. The cost per student for a van is $120/10 = $12.00. Clearly, the least-cost option makes exclusive use of vans. However, vans require more chaperones.

For 480 students, using vans alone would require 48 chaperones. Each bus carries as many students as 8 vans, but requires 4 fewer chaperones than 8 vans. So, we can cut the number of chaperones required by 8 if we add 2 buses to the mix and drop 16 vans. That means our plan is to rent 2 buses and 32 vans.