Answer :
Let's go through the steps to solve this linear programming problem.
1. Define the objective function:
We want to maximize [tex]\( z = 2x + 6y \)[/tex].
2. List the constraints:
[tex]\[ \begin{cases} 2x - 3y \leq 12 \\ x + y \leq 5 \\ 3x + 4y \geq 28 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
3. Analyze the feasible region:
The feasible region is defined by the intersection of the half-planes described by the constraints. However, it must also satisfy [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].
4. Critical points:
To find the optimal solution (i.e., the maximum value of [tex]\( z \)[/tex]), we typically look for critical points at the intersections of the boundary lines of the constraints within the feasible region. In this setup, the boundaries from the inequalities [tex]\( 2x - 3y = 12 \)[/tex], [tex]\( x + y = 5 \)[/tex], and [tex]\( 3x + 4y = 28 \)[/tex] intersect to form vertices of the feasible region.
5. Check feasibility of each vertex:
We would substitute the coordinates of each vertex into the constraints to determine if they lie within the feasible region. Once verified, we would then substitute these coordinates into the objective function [tex]\( z = 2x + 6y \)[/tex] to find the value of [tex]\( z \)[/tex].
6. Compare the values of [tex]\( z \)[/tex]:
The vertex that gives the highest value of [tex]\( z \)[/tex] is the point where [tex]\( z \)[/tex] is maximized, provided it lies within the feasible region.
After carefully examining the constraints and calculating the intersections of the given inequalities, it turns out that the feasible region does not contain any vertices that can produce a positive feasible solution set while satisfying all constraints together. Hence, it's determined that:
B. The maximum does not exist.
1. Define the objective function:
We want to maximize [tex]\( z = 2x + 6y \)[/tex].
2. List the constraints:
[tex]\[ \begin{cases} 2x - 3y \leq 12 \\ x + y \leq 5 \\ 3x + 4y \geq 28 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
3. Analyze the feasible region:
The feasible region is defined by the intersection of the half-planes described by the constraints. However, it must also satisfy [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].
4. Critical points:
To find the optimal solution (i.e., the maximum value of [tex]\( z \)[/tex]), we typically look for critical points at the intersections of the boundary lines of the constraints within the feasible region. In this setup, the boundaries from the inequalities [tex]\( 2x - 3y = 12 \)[/tex], [tex]\( x + y = 5 \)[/tex], and [tex]\( 3x + 4y = 28 \)[/tex] intersect to form vertices of the feasible region.
5. Check feasibility of each vertex:
We would substitute the coordinates of each vertex into the constraints to determine if they lie within the feasible region. Once verified, we would then substitute these coordinates into the objective function [tex]\( z = 2x + 6y \)[/tex] to find the value of [tex]\( z \)[/tex].
6. Compare the values of [tex]\( z \)[/tex]:
The vertex that gives the highest value of [tex]\( z \)[/tex] is the point where [tex]\( z \)[/tex] is maximized, provided it lies within the feasible region.
After carefully examining the constraints and calculating the intersections of the given inequalities, it turns out that the feasible region does not contain any vertices that can produce a positive feasible solution set while satisfying all constraints together. Hence, it's determined that:
B. The maximum does not exist.
