Linear Programming Problem Setup

We need to maximize the objective function $Z = 2x_1 + 5x_2$ subject to the following constraints:

• \( x_1 \leq 4 \)
• \( x_2 \leq 3 \)
• \( 2x_1 + 3x_2 \leq 14 \)
• \( x_1 \geq 0 \)
• \( x_2 \geq 0 \)

Feasible Region Corner Points Identification

The corner points of the feasible region are found by solving the constraint equations pairwise.

1. Intersection of \( x_1 = 0 \) and \( x_2 = 0 \): Point is (0, 0).
2. Intersection of \( x_1 = 4 \) and \( x_2 = 0 \): Point is (4, 0).
3. Intersection of \( x_1 = 0 \) and \( x_2 = 3 \): Point is (0, 3).
4. Intersection of \( x_1 = 4 \) and \( 2x_1 + 3x_2 = 14 \): Substitute \( x_1 = 4 \): \( 2(4) + 3x_2 = 14 \implies 8 + 3x_2 = 14 \implies 3x_2 = 6 \implies x_2 = 2 \). Point is (4, 2). This point satisfies \( x_1 \leq 4 \) and \( x_2 \leq 3 \).
5. Intersection of \( x_2 = 3 \) and \( 2x_1 + 3x_2 = 14 \): Substitute \( x_2 = 3 \): \( 2x_1 + 3(3) = 14 \implies 2x_1 + 9 = 14 \implies 2x_1 = 5 \implies x_1 = 2.5 \). Point is (2.5, 3). This point satisfies \( x_1 \leq 4 \) and \( x_2 \leq 3 \).
6. Intersection of \( x_1 = 4 \) and \( x_2 = 3 \): Point is (4, 3). Check feasibility: \( 2(4) + 3(3) = 8 + 9 = 17 \). Since \( 17 > 14 \), this point is not feasible.

The corner points of the feasible region are (0, 0), (4, 0), (0, 3), (4, 2), and (2.5, 3).

Objective Function Evaluation

Evaluate the objective function $Z = 2x_1 + 5x_2$ at each corner point:

• At (0, 0): $Z = 2(0) + 5(0) = 0$
• At (4, 0): $Z = 2(4) + 5(0) = 8$
• At (0, 3): $Z = 2(0) + 5(3) = 15$
• At (4, 2): $Z = 2(4) + 5(2) = 8 + 10 = 18$
• At (2.5, 3): $Z = 2(2.5) + 5(3) = 5 + 15 = 20$

Maximum Value Determination

Comparing the values of Z calculated at the corner points, the maximum value is 20.

Therefore, the optimal solution is $Z_{\max} = 20$.