Maximise Z=3x₁+5x₂+10x₃ subject to: x₁+x₂+x₃≤100 9x₁+x₂+8x₃≤95 x₁+8x₂+3x₃≤60 2x₁+7x₂+x₃≤150…

Examine the sign restrictions given at the end of the problem: x₁ ≥ 0, x₂ < 0, x₃ > 0. 1. In standard Linear Programming, all decision variables must satisfy non-negativity constraints (x₁, x₂, x₃ ≥ 0), which is violated by x₂ < 0. 2. Moreover, on the region defined by the strict inequalities x₂ < 0 and x₃ > 0, the feasible set is not closed. From the second constraint (9x₁ + x₂ + 8x₃ ≤ 95), as x₁ = 0 and x₂ → 0⁻, x₃ can approach (95/8)⁻ = 11.875⁻, causing the objective function Z = 3x₁ + 5x₂ + 10x₃ to approach a supremum of 118.75 that is never attained because x₂ = 0 is excluded by x₂ < 0 (and making x₂ arbitrarily negative allows x₃ to grow further if other constraints allow, or prevents attainment of the boundary maximum). Consequently, the problem has no optimal solution .