To solve this Linear Programming Problem graphically, we plot the constraints on the coordinate plane to identify the feasible region. The vertices of this region are where the objective function will achieve its minimum.
Step 1: Plot the Constraints
1. For the constraint \(x + 2y \geq 10\), rearrange to \(y \geq \frac{10 - x}{2}\). This indicates the region above the line \(x + 2y = 10\).
2. For the constraint \(x + y \geq 6\), rearrange to \(y \geq 6 - x\). This indicates the region above the line \(x + y = 6\).
3. For the constraint \(3x + y \geq 8\), rearrange to \(y \geq 8 - 3x\). This indicates the region above the line \(3x + y = 8\).
4. The constraints \(x \geq 0\) and \(y \geq 0\) define the first quadrant, where both \(x\) and \(y\) are non-negative.
Step 2: Graph the Lines and Identify the Feasible Region
The feasible region is the area where all inequalities overlap. Its vertices are the potential optimal solutions. We will evaluate the objective function \(Z = 3x + 5y\) at each vertex to find the minimum value of \(Z\).
Step 3: Evaluate the Objective Function
The feasible region is bounded by the lines and the first quadrant. The vertices of the feasible region are determined by the intersection points of these lines. The vertices are \( (4, 2) \), \( (5, 1) \), and \( (6, 0) \).
Step 4: Find the Minimum Value of \(Z = 3x + 5y\)
Evaluate \(Z\) at each vertex:
- At \( (4, 2) \): \( Z = 3(4) + 5(2) = 12 + 10 = 22 \).
- At \( (5, 1) \): \( Z = 3(5) + 5(1) = 15 + 5 = 20 \).
- At \( (6, 0) \): \( Z = 3(6) + 5(0) = 18 + 0 = 18 \).
The minimum value of \(Z\) is \( 18 \), occurring at the point \( (6, 0) \).