Answer:
4 SUV, 6 vans
Step-by-step explanation:
Let x represent the number of SUV rented, and let y represent the number of vans to be rented.
Each SUV accommodates 4 children while Each van accommodates 4 children. Since there are at most 40 children who are attending, hence:
4x + 4y ≤ 40 (1)
Also, Each SUV accommodates 3 adults while Each van accommodates 4 adults. Since there are at most 36 adults who are attending, hence:
3x + 4y ≥ 36 (2)
The system constraints are:
4x + 4y ≤ 40 (1)
3x + 4y ≥ 36 (2)
x, y ≥ 0
b) The SUV cost $85 to rent and the van cost $95 to rent. Therefore the cost equation is:
Cost = 85x + 95y
From the graph, the solution to the constraints are (0, 10) and (4, 6).
At (0, 10); Cost = 85(0) + 95(10) = $950
At (4, 6); Cost = 85(4) + 95(6) = $910
Therefore the minimum cost is at (4,6). That is only 4 SUV and 6 vans should be rented