Question

name:
mf0205

1. three teams are travelling to a basketball tournament in cars and minivans.

• each team has no more than 2 coaches and 14 athletes.
• each car can take 4 team members, and each minivan can take 6 team members.
• no more than 4 minivans and 12 cars are available.

the school wants to know what combinations will result in the minimum number of vehicles being used.
a) create a set of inequalities to model the situation, then graph them.
b) determine the minimum number of vehicles used represented by the model.

Explanation:

Part (a)

Step 1: Define Variables

Let \( x \) be the number of cars and \( y \) be the number of minivans.

Step 2: Total Team Members per Team

Each team has at most \( 2 + 14 = 16 \) members. There are 3 teams, so total members: \( 3\times16 = 48 \). The vehicle capacity: \( 4x + 6y \geq 48 \) (since they need to carry all 48 members). Simplify: \( 2x + 3y \geq 24 \).

Step 3: Vehicle Availability

Cars: \( x \leq 12 \) (no more than 12 cars), Minivans: \( y \leq 4 \) (no more than 4 minivans). Also, \( x \geq 0 \), \( y \geq 0 \) (non - negative number of vehicles).

Step 4: Graphing Inequalities

• For \( 2x + 3y \geq 24 \): First, graph the line \( 2x + 3y = 24 \). When \( x = 0 \), \( y = 8 \); when \( y = 0 \), \( x = 12 \). Since it's \( \geq \), shade above the line.
• For \( x \leq 12 \): Graph the vertical line \( x = 12 \) and shade to the left.
• For \( y \leq 4 \): Graph the horizontal line \( y = 4 \) and shade below.
• For \( x\geq0,y\geq0 \): Shade in the first quadrant.

Step 1: Objective Function

We want to minimize the number of vehicles \( V=x + y \).

Step 2: Analyze Constraints

We know \( y\leq4 \). Let's substitute \( y = 4 \) into \( 2x+3y\geq24 \):
\( 2x+3\times4\geq24 \)
\( 2x + 12\geq24 \)
\( 2x\geq12 \)
\( x\geq6 \)

Step 3: Calculate \( V=x + y \)

If \( y = 4 \) and \( x = 6 \), then \( V=6 + 4=10 \). Let's check other possible values of \( y \) (since \( y\leq4 \), \( y = 0,1,2,3,4 \)):

• If \( y = 3 \), then \( 2x+9\geq24\Rightarrow2x\geq15\Rightarrow x\geq7.5\), so \( x = 8 \), \( V=11 \).
• If \( y = 4 \), \( x = 6 \) gives \( V = 10 \). If we try \( y=4 \) and \( x = 6 \), \( 4\times6+6\times4=24 + 24 = 48 \) (which is exactly the number of team members).

Answer:

(for part a):
The inequalities are \( 2x + 3y \geq 24 \), \( x \leq 12 \), \( y \leq 4 \), \( x\geq0 \), \( y\geq0 \). The graph is as described (shaded region satisfying all inequalities).

Part (b)