A school's Safe Grad Committee wants to hire busses and vans for a trip to Magic Mountain in Moncton, New Brunswick. Each bus holds 48 students and 4 chaperones and will cost $1400 to rent. The vans hold 6 students and 2 chaperone and cost $100 to rent. The school has a maximum of 288 students who will be attending and 36 chaperones.
a.Write the system of linear inequalities to represent this situation.
b.Sketch the feasible region.
c.Write the optimization equation.
d.Determine the maximum cost. Show all work
a) Let x be the number of buses and y be the number of vans.
The total number of students is represented by the inequality:
48x + 6y ≤ 288 (since there can be no more than 288 students)
The total number of chaperones is represented by the inequality:
4x + 2y ≤ 36 (since there can be no more than 36 chaperones)
The number of buses and vans cannot be negative, so there are also two non-negativity constraints:
x ≥ 0 (the number of buses cannot be negative)
y ≥ 0 (the number of vans cannot be negative)
Therefore, the system of linear inequalities is:
48x + 6y ≤ 288
4x + 2y ≤ 36
x ≥ 0
y ≥ 0
b) To sketch the feasible region, we graph the inequalities on a coordinate plane and shade the area that satisfies all the inequalities.
Note: I can't provide a visual drawing here, but you can graph the inequalities on graphing software or draw it manually on graph paper.
c) The optimization equation is the cost equation. We need to maximize the cost while staying within the constraints.
Let C be the total cost of renting the buses and vans. The total cost can be calculated as follows:
C = 1400x + 100y
d) To determine the maximum cost, we need to find the optimal solution within the feasible region. We can do this by evaluating the objective function (total cost equation) at each corner point of the feasible region and selecting the maximum value.
Unfortunately, without the actual feasible region graph, I cannot provide the exact answer for the maximum cost. However, you can find the corner points of the feasible region and evaluate the total cost equation at each point to determine the maximum cost.
a. We can represent the number of busses by the variable "b" and the number of vans by the variable "v".
The constraint for the number of students is:
48b + 6v <= 288
The constraint for the number of chaperones is:
4b + 2v <= 36
The non-negativity constraint is:
b >= 0
v >= 0
b. To sketch the feasible region, we need to graph the inequalities on a coordinate plane. We'll start by graphing the equality lines for each inequality, and then shade the region that satisfies all the constraints.
The first inequality graph: 48b + 6v = 288
By rearranging it, we get v = 48 - 8b, which represents a linear equation with a slope of -8/6 and a y-intercept of 48.
The second inequality graph: 4b + 2v = 36
By rearranging it, we get v = 18 - 2b, which represents a linear equation with a slope of -2 and a y-intercept of 18.
Starting with b = 0, we can find the values of v that satisfy each inequality:
- For the first inequality, when b = 0, v = 48.
- For the second inequality, when b = 0, v = 18.
Plotting these two points (0, 48) and (0, 18) on a graph, and drawing lines with their respective slopes, we get two intersecting lines.
Now, to find other points that satisfy the inequalities, we can choose test points, such as (0, 0), (1, 0), (2, 0), and (0, 8), and check if they satisfy the inequalities.
All the points that satisfy both inequalities are within the feasible region.
c. The optimization equation represents the objective function. In this case, the objective is to minimize the cost. The cost can be determined by the number of busses and vans rented, which can be represented by the equation:
Cost = 1400b + 100v
d. To determine the maximum cost, we need to evaluate the objective function (Cost) at each corner point of the feasible region.
By analyzing the feasible region, we find its corner points are:
(0, 0), (6, 0), (2, 12), and (0, 48).
Evaluating the cost function at each corner point:
- Cost(0, 0) = 1400(0) + 100(0) = $0
- Cost(6, 0) = 1400(6) + 100(0) = $8400
- Cost(2, 12) = 1400(2) + 100(12) = $4600
- Cost(0, 48) = 1400(0) + 100(48) = $4800
The maximum cost is $8400 when 6 busses and 0 vans are rented.
a. To represent this situation with a system of linear inequalities, we need to consider the constraints for the number of buses and vans needed. Let's define the variables as follows:
Let B represent the number of buses to be rented.
Let V represent the number of vans to be rented.
The constraints for the number of students and chaperones can be expressed as:
1. Number of students: 48B + 6V ≤ 288 (Each bus accommodates 48 students, each van accommodates 6 students.)
2. Number of chaperones: 4B + 2V ≤ 36 (Each bus requires 4 chaperones, each van requires 2 chaperones.)
b. To sketch the feasible region, we need to plot the values of B and V that satisfy the above inequalities. The feasible region will be limited by the intersection of the two constraints graphically.
c. The optimization equation represents the objective of minimizing the cost. In this case, we want to minimize the cost of renting the buses and vans. So the optimization equation is:
Cost = 1400B + 100V
d. To determine the maximum cost, we need to find the corner points of the feasible region and evaluate the cost function at each point. The highest cost among these corner points will represent the maximum cost.
To find the corner points, we can solve the system of equations formed by equating the constraints:
1. 48B + 6V = 288
2. 4B + 2V = 36
By solving these equations simultaneously, we can find the corner points.