Question

Solve the following Linear Programming Problem graphically:
Minimise \( Z = 3x + 5y \)
subject to the constraints: \[ x + 2y \geq 10, \quad x + y \geq 6, \quad 3x + y \geq 8, \quad x, y \geq 0. \]

Hint:

When solving Linear Programming Problems graphically, always start by plotting the constraints, find the feasible region, and then evaluate the objective function at each vertex of the feasible region to find the optimal solution.

Answer 1

To solve this Linear Programming Problem graphically, we plot the constraints on the coordinate plane and then find the feasible region formed by these constraints. The vertices of the feasible region will provide the points at which the objective function will attain its minimum. Step 1: Plot the Constraints 1. For the constraint \(x + 2y \geq 10\), rearrange it as: \[ y \geq \frac{10 - x}{2}. \] This represents the region above the line \(x + 2y = 10\). 2. For the constraint \(x + y \geq 6\), rearrange it as: \[ y \geq 6 - x. \] This represents the region above the line \(x + y = 6\). 3. For the constraint \(3x + y \geq 8\), rearrange it as: \[ y \geq 8 - 3x. \] This represents the region above the line \(3x + y = 8\). 4. The last two constraints \(x \geq 0\) and \(y \geq 0\) represent the first quadrant of the coordinate plane, i.e., the region 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 the inequalities overlap. The vertices of this feasible region are the potential solutions. We will evaluate the objective function \(Z = 3x + 5y\) at each of these vertices 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. Let's denote the points of intersection of these lines. From the graph:
- The vertices of the feasible region are \( (4, 2) \), \( (5, 1) \), and \( (6, 0) \).
Step 4: Find the Minimum Value of \(Z = 3x + 5y\) At the vertices:
- At \( (4, 2) \), \( Z = 3(4) + 5(2) = 12 + 10 = 22 \).
- At \( (5, 1) \), \( Z = 3(5) + 5(1) = 15 + 5 = 20 \).
- At \( (6, 0) \), \( Z = 3(6) + 5(0) = 18 + 0 = 18 \).
Thus, the minimum value of \(Z\) occurs at the point \( (6, 0) \), and the minimum value of \(Z\) is \( 18 \).