Let's carefully analyze and solve the given problem, step by step.
The objective is to maximize [tex]\( z = 5x + 3y \)[/tex].
Given the constraints:
1. [tex]\(5x + 5y \leq 9\)[/tex]
2. [tex]\(x + y \leq 24\)[/tex]
3. [tex]\(3x + 3y \leq 3\)[/tex]
We will introduce slack variables to transform inequalities into equalities for linear programming:
1. [tex]\(5x + 5y + s_1 = 9\)[/tex]
2. [tex]\(x + y + s_2 = 24\)[/tex]
3. [tex]\(3x + 3y + s_3 = 3\)[/tex]
Where [tex]\(s_1, s_2,\)[/tex] and [tex]\(s_3\)[/tex] are non-negative slack variables.
Now, the linear programming problem with slack variables becomes:
Maximize [tex]\( z = 5x + 3y \)[/tex]
Subject to:
[tex]\[ 5x + 5y + s_1 = 9 \][/tex]
[tex]\[ x + y + s_2 = 24 \][/tex]
[tex]\[ 3x + 3y + s_3 = 3 \][/tex]
[tex]\[ s_1, s_2, s_3 \geq 0 \][/tex]
To solve this, we need to find the basic feasible solutions by setting two variables to zero and solving for the rest. Given the constraints, it becomes obvious that there is no valid solution, since the constraints are contradictory.
Let’s quickly state the contradictions:
1. If [tex]\(5x + 5y + s_1 = 9\)[/tex]:
- Dividing by 5, [tex]\(x + y + \frac{s_1}{5} = 1.8\)[/tex]
2. If [tex]\(x + y + s_2 = 24\)[/tex]
- Directly shows that [tex]\(x + y\)[/tex] should be less than or equal to 24
3. If [tex]\(3x + 3y + s_3 = 3\)[/tex]
- Dividing by 3, [tex]\(x + y + \frac{s_3}{3} = 1\)[/tex]
Comparing them:
- From conditions, [tex]\(x + y \leq 1\)[/tex] contradicts [tex]\(x + y \leq 24\)[/tex] and also [tex]\(x + y \leq 1.8\)[/tex]
Therefore, these inequalities do not make sense together, indicating a contradiction in provided constraints. Thus, there are no feasible solutions for [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
To conclude, given the constraints, there is no valid solution to the given linear programming problem. Realistically, the linear programming formulation would require feasible and non-contradictory constraints for further solving.
