Answer:
a) MAX--> PC (R,P) = 0,3R+ 0,5P
b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000
c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.
d) <u>Binding Constaints</u>:
1. 0.3 R+0.6 P ≤ 18.000
2. R+P ≤ 50.000
3. P ≤ 20.000
4. R ≥ 0
5. P ≥ 0
Step-by-step explanation:
I will solve it using the graphic method:
First, we have to define the variables:
R : Regular Gasoline
P: Premium Gasoline
We also call:
PC: Profit contributions
A: Grade A crude oil
• R--> PC: $0,3 --> 0,3 A
• P--> PC: $0,5 --> 0,6 A
So the ecuation to maximize is:
MAX--> PC (R,P) = 0,3R+ 0,5P
The restrictions would be:
1. 18.000 A availabe (R=0,3 A ; P 0,6 A)
2. 50.000 capacity
3. Demand of P: No more than 20.000
4. Both P and R 0 or more.
Translated to formulas:
Answer d)
1. 0.3 R+0.6 P ≤ 18.000
2. R+P ≤ 50.000
3. P ≤ 20.000
4. R ≥ 0
5. P ≥ 0
To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.
So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)
V:(R;P)
• V1: (0;0)
• V2: (0; 20.000)
• V3: (20.000;20.000)
• V4: (40.000; 10.000)
• V5:(50.000;0)
We check each one in the profit ecuation:
MAX--> PC (R,P) = 0,3R+ 0,5P
• V1: 0
• V2: 10.000
• V3: 16.000
• V4: 17.000
• V5: 15.000
As we can see, the optimal solution is
V4: 40.000 units of regular and 10.000 of premium.
To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.
