Question

Consider the following Linear Programming Problem P:
Minimize \(3x_1 + 4x_2\)
subject to \[ x_1 - x_2 \le 1, \] \[ x_1 + x_2 \ge 3, \] \[ x_1 \ge 0, x_2 \ge 0. \] The optimal value of the problem P is ..................

Hint:

For minimization problems with an unbounded feasible region, if the objective function's coefficients are all non-negative, the optimal solution will always exist and occur at a corner point. There is no need to test points at infinity.

Answer 1

Step 1: Understanding the Concept:
This is a linear programming problem. For a problem with two variables, we can solve it graphically. We need to identify the feasible region defined by the constraints, find the coordinates of the corner points (vertices) of this region, and then evaluate the objective function at these points to find the minimum value.
Step 3: Detailed Explanation:
Step 3.1: Graph the feasible region.
The constraints are: 1. \(x_1 - x_2 \le 1 \implies x_2 \ge x_1 - 1\). The boundary is the line \(x_2 = x_1 - 1\). The feasible region is above this line. 2. \(x_1 + x_2 \ge 3 \implies x_2 \ge -x_1 + 3\). The boundary is the line \(x_2 = -x_1 + 3\). The feasible region is above this line. 3. \(x_1 \ge 0\) and \(x_2 \ge 0\). This restricts the region to the first quadrant. The feasible region is the area in the first quadrant that is above both lines. This region is unbounded from the top.
Step 3.2: Find the corner points.
The corner points are the vertices of the feasible region.
- Point A: Intersection of \(x_1 = 0\) and \(x_1 + x_2 = 3\).
Substituting \(x_1=0\) gives \(0 + x_2 = 3 \implies x_2 = 3\).
So, Point A is \((0, 3)\).
Check other constraints: \(0-3 = -3 \le 1\) (True).
- Point B: Intersection of the two lines \(x_1 - x_2 = 1\) and \(x_1 + x_2 = 3\). Adding the two equations: \((x_1 - x_2) + (x_1 + x_2) = 1 + 3 \implies 2x_1 = 4 \implies x_1 = 2\).
Substituting \(x_1=2\) into \(x_1 + x_2 = 3\) gives \(2 + x_2 = 3 \implies x_2 = 1\). So, Point B is \((2, 1)\).
The feasible region has two corner points: (0, 3) and (2, 1).
Step 3.3: Evaluate the objective function at the corner points.
The objective function to minimize is \(Z = 3x_1 + 4x_2\).
- At Point A (0, 3): \(Z = 3(0) + 4(3) = 12\).
- At Point B (2, 1): \(Z = 3(2) + 4(1) = 6 + 4 = 10\).
Since the feasible region is unbounded, we must verify that the minimum is not unbounded. The objective function coefficients are positive, so as \(x_1\) or \(x_2\) increase, Z increases. Thus, the minimum must occur at one of the corner points.
Step 4: Final Answer:
Comparing the values, the minimum value is 10, which occurs at the point (2, 1).
Step 5: Why This is Correct:
The graphical method correctly identifies the feasible region and its vertices. The evaluation of the objective function at these vertices shows that the minimum value is 10. Since the objective function increases as we move away from the origin within the feasible region, this minimum value is the optimal solution.