Understanding the Linear Programming Problem (LPP)

The problem presented is a Linear Programming Problem (LPP). Our goal is to find the values of variables x₁ and x₂ that maximize a given objective function, while satisfying a set of linear inequalities called constraints. The non-negativity constraints (x₁ ≥ 0, x₂ ≥ 0) mean that the variables cannot take negative values, which is typical in many real-world applications like production or resource allocation.

The specific LPP is:

Objective Function:

$$ \text{Maximize } z = x_1 + x_2 $$

Subject to the Constraints:

• $$ x_1 + 2x_2 \le 2000 $$ (Constraint 1)
• $$ x_1 + x_2 \le 1500 $$ (Constraint 2)
• $$ x_2 \le 600 $$ (Constraint 3)
• $$ x_1 \ge 0, x_2 \ge 0 $$ (Non-negativity Constraints)

Solving the Linear Programming Problem Graphically

Since this LPP involves only two variables (x₁ and x₂), we can solve it using the graphical method. This involves the following steps:

1. Convert each inequality constraint into a linear equation to graph the boundary lines.
2. Graph these lines on a 2D plane (with x₁ on the horizontal axis and x₂ on the vertical axis).
3. Determine the feasible region, which is the area that satisfies all the constraints simultaneously, including the non-negativity constraints (the first quadrant).
4. Identify the corner points (vertices) of the feasible region.
5. Evaluate the objective function at each corner point.
6. The corner point that gives the maximum value of the objective function is the optimal solution.

Graphing the Constraints and Finding the Feasible Region

Let's convert the inequalities to equations:

• L1: $$ x_1 + 2x_2 = 2000 $$
• L2: $$ x_1 + x_2 = 1500 $$
• L3: $$ x_2 = 600 $$

We also have $$ x_1 = 0 $$ (y-axis) and $$ x_2 = 0 $$ (x-axis) from non-negativity.

We need to find the corner points of the feasible region defined by these lines and the non-negativity constraints.

Identifying the Corner Points

The corner points are the intersections of the boundary lines that form the vertices of the feasible region:

• Point A: Intersection of $$ x_1 = 0 $$ and $$ x_2 = 0 $$. This is the origin (0, 0).
• Point B: Intersection of $$ x_1 = 0 $$ and $$ x_2 = 600 $$. This point is (0, 600). Let's check if it satisfies other constraints:
  • $$ 0 + 2(600) = 1200 \le 2000 $$ (Satisfied)
  • $$ 0 + 600 = 600 \le 1500 $$ (Satisfied)
  • $$ 0 \ge 0, 600 \ge 0 $$ (Satisfied)

  Point B(0, 600) is a valid corner point.

• Point C: Intersection of $$ x_2 = 600 $$ and $$ x_1 + 2x_2 = 2000 $$. Substitute $$ x_2 = 600 $$ into the second equation:

  $$ x_1 + 2(600) = 2000 $$

  $$ x_1 + 1200 = 2000 $$

  $$ x_1 = 800 $$

  This point is (800, 600). Let's check if it satisfies other constraints:

  • $$ 800 + 600 = 1400 \le 1500 $$ (Satisfied)
  • $$ 800 \ge 0, 600 \ge 0 $$ (Satisfied)

  Point C(800, 600) is a valid corner point.

• Point D: Intersection of $$ x_1 + 2x_2 = 2000 $$ and $$ x_1 + x_2 = 1500 $$. Subtract the second equation from the first:

  $$ (x_1 + 2x_2) - (x_1 + x_2) = 2000 - 1500 $$

  $$ x_2 = 500 $$

  Substitute $$ x_2 = 500 $$ into $$ x_1 + x_2 = 1500 $$:

  $$ x_1 + 500 = 1500 $$

  $$ x_1 = 1000 $$

  This point is (1000, 500). Let's check if it satisfies other constraints:

  • $$ 500 \le 600 $$ (Satisfied)
  • $$ 1000 \ge 0, 500 \ge 0 $$ (Satisfied)

  Point D(1000, 500) is a valid corner point.

• Point E: Intersection of $$ x_1 + x_2 = 1500 $$ and $$ x_2 = 0 $$. Substitute $$ x_2 = 0 $$ into the first equation:

  $$ x_1 + 0 = 1500 $$

  $$ x_1 = 1500 $$

  This point is (1500, 0). Let's check if it satisfies other constraints:

  • $$ 1500 + 2(0) = 1500 \le 2000 $$ (Satisfied)
  • $$ 0 \le 600 $$ (Satisfied)
  • $$ 1500 \ge 0, 0 \ge 0 $$ (Satisfied)

  Point E(1500, 0) is a valid corner point.

The corner points of the feasible region are (0,0), (0,600), (800,600), (1000,500), and (1500,0).

Evaluating Objective Function at Corner Points

Now, we evaluate the objective function $$ z = x_1 + x_2 $$ at each corner point:

Identifying the Optimal Solution

Comparing the values of z calculated at each corner point, the maximum value of z is 1500. This maximum occurs at two corner points: (1000, 500) and (1500, 0).

The options provided are potential solutions:

• Option 1: x₁=750, x₂=750, z=1500. Check feasibility: 750 + 2(750) = 2250. This exceeds 2000, so it violates Constraint 1. This point is not in the feasible region.
• Option 2: x₁=500, x₂=1000, z=1500. Check feasibility: 500 + 2(1000) = 2500. This exceeds 2000, so it violates Constraint 1. This point is not in the feasible region.
• Option 3: x₁=1000, x₂=500, z=1500. Check feasibility:
  • $$ 1000 + 2(500) = 2000 \le 2000 $$ (Satisfied)
  • $$ 1000 + 500 = 1500 \le 1500 $$ (Satisfied)
  • $$ 500 \le 600 $$ (Satisfied)
  • $$ 1000 \ge 0, 500 \ge 0 $$ (Satisfied)

  This point is in the feasible region, and the objective function value is $$ 1000 + 500 = 1500 $$. This matches the maximum value found at a corner point.

• Option 4: x₁=900, x₂=600, z=1500. Check feasibility: 900 + 2(600) = 900 + 1200 = 2100. This exceeds 2000, so it violates Constraint 1. This point is not in the feasible region.

Among the given options, only x₁=1000, x₂=500 is a feasible solution that yields a z value of 1500, which is the maximum possible value.

Revision Table: Linear Programming Concepts

Additional Information on LPP Solutions

It's important to know that an LPP doesn't always have a unique optimal solution. Possible scenarios for LPP solutions include:

• Unique Optimal Solution: The objective function is maximized or minimized at exactly one corner point of the feasible region.
• Multiple Optimal Solutions: The objective function is maximized or minimized at two or more corner points. In this case, any point on the line segment connecting these corner points is also an optimal solution. In our example, both (1000, 500) and (1500, 0) yield the maximum z=1500. So, there are multiple optimal solutions along the line segment connecting these points, subject to remaining within the feasible region.
• Unbounded Solution: The feasible region is not bounded in the direction of optimization (e.g., maximization), meaning the objective function can increase indefinitely.
• Infeasible Solution: There is no feasible region because the constraints are contradictory.

For LPPs with more than two variables, the graphical method is not applicable, and other techniques like the Simplex method are used.