The production planner for a private label soft drink maker is planning the production of two soft drinks: root beer (R)and sassafras soda (S).Two resources are constrained: production time (T),of which she has at most 12 hours per day; and carbonated water (W),of which she can get at most 1500 gallons per day.A case of root beer requires 2 minutes of time and 5 gallons of water to produce,while a case of sassafras soda requires 3 minutes of time and 5 gallons of water.Profits for the root beer are $6.00 per case,and profits for the sassafras soda are $4.00 per case.Which of the following is not a feasible production combination?
A)0 R & 0 S
B)0 R & 240 S
C)180 R & 120 S
D)300 R & 0 S
E)180 R & 240 S

Question

Quizplus · Accepted Answer

To determine the feasible production combinations, we need to consider the constraints:
1. Production time constraint: At most 12 hours per day, which is equal to 720 minutes per day.
2. Carbonated water constraint: At most 1500 gallons per day.

Let's represent the production of root beer as R and the production of sassafras soda as S. The constraints can be represented as follows:
2R + 3S ≤ 720 (production time constraint)
5R + 5S ≤ 1500 (carbonated water constraint)

We want to find the combinations of R and S that satisfy these constraints. We also want to maximize the profit.

Let's write the profit equation:
Profit = 6R + 4S

We can convert this profit equation into a constraint by setting a minimum profit level, say $0:
6R + 4S ≥ 0

Now we have the following system of inequalities:
2R + 3S ≤ 720
5R + 5S ≤ 1500
6R + 4S ≥ 0

We can solve this system of inequalities graphically or algebraically. One way to solve it algebraically is by using linear programming techniques. Using a computer program, we can find the optimal solution as follows:

Answer: 240 S
 
The optimal solution is to produce 240 cases of sassafras soda and 0 cases of root beer. This will result in a profit of $960 per day, which is the maximum achievable profit given the constraints.

Here's how to get this answer. We can rewrite the system of inequalities as follows:
2R + 3S ≤ 720
R + S ≤ 300 (dividing the second constraint by 5)
6R + 4S ≥ 0

Now we can graph these inequalities and find the feasible region. The feasible region is the shaded area in the graph below:

```
                   2R + 3S ≤ 720 
                   R + S  ≤ 300 
                   6R + 4S ≥   0 
```

We want to maximize the profit equation 6R + 4S subject to the constraints. The optimal solution occurs at the corner point of the feasible region that has the highest profit. The corner points are (0,0), (0,300), (180,120), (240,0), and (150,150). We can evaluate the profit equation at each corner point to determine the highest profit:

(0,0): Profit = 6(0) + 4(0) = $0
(0,300): Profit = 6(0) + 4(300) = $1200
(180,120): Profit = 6(180) + 4(120) = $1680
(240,0): Profit = 6(240) + 4(0) = $1440
(150,150): Profit = 6(150) + 4(150) = $1800

The highest profit is obtained at the point (150,150), but this point is not feasible because it violates the carbonated water constraint. So the optimal solution is to produce 240 cases of sassafras soda and 0 cases of root beer, which gives a profit of $960.

Therefore, the not feasible production combination is E) 180 R & 240 S, as it violates the production time constraint.

The Production Planner for a Private Label Soft Drink Maker