I am writing to you because I need you to explain the solution to this exercise. I am very sorry, I have not found anything that will help me solve it. I know that in this forum there are teachers with a fairly advanced level and I trust that I can find help.
For the problem: Max. f(x) subject to $x\in{}X, where: \\X = \{ x\in{}\Bbb R^3 | 4x_1 + 2x_2 + x_3 \leq 100, x_1 + x_2 + x_3 \leq 600, x_1, x_2, x_3 \geq 0\}\\ f = (f_1, f_2), f_1(x) = 4x_1 + 3x_2 + x_3, f_2(x) = x_2 + x_3$
- Determine the set of efficient extreme points using the parameterization method. 2) Determine the set of efficient points.
This exercise is optimization.
To find the efficient extreme points, we first need to find the extreme points of the set X. To do this, we equate the constraints to the corresponding equations:
4x1 + 2x2 + x3 = 100x1 + x2 + x3 = 600
Solving this system of equations, we obtain the extreme points of the set X:
x1 = 200, x2 = 200, x3 = 200x1 = 0, x2 = 0, x3 = 600x1 = 0, x2 = 300, x3 = 300
Now we are going to parameterize the objective functions f1 and f2 with respect to a parameter t:
f1(x) = 4x1 + 3x2 + x3f2(x) = x2 + x3
We substitute the values of the extreme points into the objective functions:
For the point (200, 200, 200):f1(t) = 4(200) + 3(200) + (200) = 1400f2(t) = (200) + (200) = 400
For the point (0, 0, 600):f1(t) = 4(0) + 3(0) + (600) = 600f2(t) = (0) + (600) = 600
For the point (0, 300, 300):f1(t) = 4(0) + 3(300) + (300) =1200f2(t) =(300)+(300)=600
Therefore, the efficient endpoints are:(200,200,200)(0,300,300)
Step 2: Determine the set of efficient points.
To determine the set of efficient points, we compare the solutions obtained in the previous step with respect to efficiency. In this case, we see that the point (200,200,200) is more efficient than the point (0,300,300), since it has a larger value in both objective functions.
Therefore, the set of efficient points is given by:{(200,200,200)}