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The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 81students, requires 8 chaperones, and costs $1,100 to rent. Each van can transport 9 students, requires 1 chaperone,and costs $90 to rent. Since there are 891 students in the senior class that may be eligible to go on the trip, the officersmust plan to accommodate at least 891 students. Since only 96 parents have volunteered to serve as chaperones, theofficers must plan to use at most 96 chaperones. How many vehicles of each type should the officers rent in order tominimize the transportation costs? What are the minimal transportation costs?The officers should rentbuses andvans to minimize the transportation costs.

Respuesta :

Given:

• Number of students each bus can transport = 81

,

• Number of chaperons each bus can transport = 8

,

• Cost to rent one bus = $1100

,

• Number of students each van can transport = 9

,

• Number of chaperons each bus can transport = 1

,

• Cost to rent one van = $90

,

• Number of eligible students = 891.

,

• Number of parents that volunteered = 96

Let's find the number of vehicles of each the officers should rent in order to minimize transportation costs.

Where:

x is the number of buses

y is the number of vans

Z is the minimal transportation cost.

The objective is to minimize the total cost:

Z = 1100x + 90y

From the given situation, we have the system of inequalities:

81x + 9y ≥ 891

8x + 1y ≤ 96

Let's solve the system of inequalities:

Rewrite the second inequality for y:

y ≥ 96 - 8x

Subsitute 96 - 8x for y in the first inequality:

[tex]\begin{gathered} 81x+9y\ge891 \\ \\ 81x+9(96-8x)\ge891 \\ \\ 81x+864-72x\ge891 \\ \\ 81x-72x+864\ge891 \\ \\ 9x\ge891-864 \\ \\ 9x\ge27 \\ \\ x\ge\frac{27}{9} \\ \\ x\ge3 \end{gathered}[/tex]

Now, plug in 3 for x in (y ≥ 96 - 8x):

[tex]\begin{gathered} y\ge96-8(3) \\ \\ y\ge96-24 \\ \\ y\ge72 \end{gathered}[/tex]

Substitute 72 for y and 3 for x in the first equation:

[tex]\begin{gathered} Z=1100x+90y \\ \\ Z=1100(3)+90(72) \\ \\ Z=3300+6480 \\ \\ Z=9780 \end{gathered}[/tex]

Thus, we have the solutions:

x = 3, y = 72, Z = 9780

Therefore, the officers should rent 3 buses and 72 vans to minimize the transportation cost.

The minimal transportation cost is $9780

ANSWER:

The officers should rent 3 buses and 72 vans to minimize the transportation cost.

Minimal transportation cost = $9780

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