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.
Let's define the variables as follows:
Let x be the number of buses rented and y be the number of vans rented.
a. The system of linear inequalities can be represented as:
Inequality 1: 48x + 6y ≤ 288 (total students cannot exceed 288)
Inequality 2: 4x + 2y ≤ 36 (total chaperones cannot exceed 36)
b. To sketch the feasible region, we plot the graph for each inequality and shade the region that satisfies both conditions:
For Inequality 1: 48x + 6y ≤ 288
If we set x = 0, we get 6y ≤ 288, which gives y ≤ 48.
If we set y = 0, we get 48x ≤ 288, which gives x ≤ 6.
Plotting these points and connecting them, we get a line. The region below this line is feasible.
For Inequality 2: 4x + 2y ≤ 36
If we set x = 0, we get 2y ≤ 36, which gives y ≤ 18.
If we set y = 0, we get 4x ≤ 36, which gives x ≤ 9.
Plotting these points and connecting them, we get another line. The region below this line is feasible.
The feasible region is the overlapping shaded region below both lines.
c. The optimization equation represents the cost of renting the buses and vans:
Cost = (Cost per bus * number of buses) + (Cost per van * number of vans)
Cost = (1400 * x) + (100 * y)
d. To determine the maximum cost, we need to find the corner point of the feasible region that maximizes the cost. We can substitute the coordinates of each corner point into the optimization equation and calculate the cost. The highest cost among these corner points will be the maximum cost.
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a. To represent this situation with a system of linear inequalities, we need to consider the constraints related to the number of buses and vans needed, as well as the number of students and chaperones attending the trip.
Let:
x = the number of buses to rent
y = the number of vans to rent
The constraints are:
1. The total number of students should not exceed 288:
48x + 6y ≤ 288
2. The total number of chaperones should not exceed 36:
4x + 2y ≤ 36
3. The number of buses and vans should be positive:
x ≥ 0
y ≥ 0
b. To sketch the feasible region, we graph the constraints on an x-y plane. The feasible region is the area where all the constraints are satisfied. Since the constraints are inequalities, the feasible region will be a bounded area.
The line 48x + 6y = 288 can be rewritten as y = (288 - 48x)/6, which simplifies to y = 48 - 8x.
Similarly, the line 4x + 2y = 36 can be rewritten as y = (36 - 4x)/2, which simplifies to y = 18 - 2x.
Plotting these lines on a graph, along with the constraints x ≥ 0 and y ≥ 0, we get the feasible region.
c. The optimization equation represents the objective of maximizing the cost. In this case, the equation would be:
Cost = 1400x + 100y
d. To determine the maximum cost, we need to find the coordinates of the vertices of the feasible region and calculate the cost for each vertex. The vertex with the highest cost will be the maximum.
Let's find the vertices:
1. Intersection of y = 48 - 8x and x-axis:
48 - 8x = 0
8x = 48
x = 6, y = 0
2. Intersection of y = 48 - 8x and y = 18 - 2x:
48 - 8x = 18 - 2x
-6x = -30
x = 5, y = 8
3. Intersection of y = 18 - 2x and y-axis:
18 - 2x = 0
2x = 18
x = 9, y = 0
Now, calculate the cost for each vertex:
Cost1 = 1400(6) + 100(0) = $8400
Cost2 = 1400(5) + 100(8) = $8500
Cost3 = 1400(9) + 100(0) = $12600
The maximum cost is $12,600.