You are in charge of renting SUVs and vans for your family reunion, and you want to minimize costs. You know that at most 40 children are attending and that at least 36 adults are attending. Each SUV accommodates 4 children and 3 adults, and costs $85 to rent. Each van accommodates 4 children and 4 adults, and costs $95 to rent. A. Set up the linear programming problem by writing the objective function as well as the system of constraints. B. How many SUVs and vans should be rented to minimize cost

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

constant of proportionality = 3 and the equation is c = 3p.

step-by-step explanation:

c = kp   where c = amount of caffeine in mg., p = oz of pepsi and k is the constant of proportionality.

substituting the values given:

24 = 8k

k = 3.

the same value of k will be obtained if we substitute the other given values.