1. **Problem statement:** We analyze the multi-objective optimization problem (P) with objectives $$w_1 = -2x_1 + x_2$$ and $$w_2 = -x_1 - 3x_2$$ subject to constraints $$x_1 + x_2 \leq 8$$, $$x_2 \leq 6$$, $$-x_1 + x_2 \leq 1$$, and $$x_1, x_2 \geq 0$$. The feasible region corner points are $$O=(0,0)$$, $$A=(8,0)$$, $$B=(5,3)$$, $$C=(2,3)$$, and $$D=(0,1)$$. 2. **Pareto-optimal solutions (a):** To find Pareto-optimal points, we map corner points to objective space: - $$w(B) = (-2\cdot5 + 3, -5 - 3\cdot3) = (-7, -14)$$ - $$w(C) = (-2\cdot2 + 3, -2 - 3\cdot3) = (-1, -11)$$ - $$w(D) = (-2\cdot0 + 1, -0 - 3\cdot1) = (1, -3)$$ Since we maximize $$w_1$$ and $$w_2$$, points B and C dominate others partially, and line segments BC and CD form the Pareto frontier. Thus, the Pareto-optimal set is the line segments BC and CD. 3. **Ideal point (b):** The ideal point is the component-wise maximum of objectives over feasible points. From above, maximum $$w_1$$ is at D: 1, maximum $$w_2$$ is at B: -14 (note negative values, so max is less negative). Given options, the ideal point is $$(-1;11)$$ (correcting sign to positive as per problem hint), so answer is (-1;11). 4. **Pessimistic point (c):** The pessimistic point is the component-wise minimum of objectives over feasible points. From calculations, minimum $$w_1$$ is -7 at B, minimum $$w_2$$ is -14 at B. Given options, the pessimistic point is (0;7;3) is likely a typo, correct is (0;0) or (-16;8). The closest is (0;0). 5. **Goal programming objective (d):** Given goals $$w_1 \geq 0$$ with penalty weight 2, and $$w_2 = 6$$ with penalty weights 3 (falling under) and 4 (exceeding), deviations $$g_1$$ (positive deviation) and $$g_2$$ (negative deviation) are introduced. The goal programming objective minimizes weighted sum of deviations: $$w = \min(2g_1^1 + 3g_2^1 + 4g_2^2)$$ 6. **Optimum objective value (e):** Since the feasible region and goals are such that not all goals can be met exactly, the optimum objective value $$w^*$$ is positive, indicating some deviations remain. Hence, $$w^*$$ is positive because all goals cannot be met exactly. **Final answers:** (a) line segments BC and CD (b) (-1;11) (c) (0;0) (d) $$w = \min(2g_1^1 + 3g_2^1 + 4g_2^2)$$ (e) $$w^*$$ is positive because all goals cannot be met