MHTCET 20 Apr 2026 Shift 2 Paper

Concept:A linear programming problem has a unique solution when the maximum value occurs at exactly one corner point of the bounded feasible region.Explanation:The given constraints are:$$x+3y \leq 6, \quad 2x+6y \leq 18, \quad x \geq 0, \quad y \geq 0$$Simplify the second constraint by dividing by $2$:$$x+3y \leq 9$$Now compare the two main constraints:$$x+3y \leq 6 \quad \text{and} \quad x+3y \leq 9$$Since $x+3y \leq 6$ is stricter, the constraint $x+3y \leq 9$ is redundant and does not change the feasible region.Thus, the effective feasible region is given by:$$x+3y \leq 6, \quad x \geq 0, \quad y \geq 0$$This is a bounded triangular region with corner points:$$(0,0), \quad (6,0), \quad (0,2)$$Evaluate $z=2x+5y$ at each corner point:At $(0,0)$: $z = 0$At $(6,0)$: $z = 12$At $(0,2)$: $z = 10$The maximum value is $12$, which occurs only at $(6,0)$.Therefore, the LPP has exactly one optimal solution.Answer:Option A: Unique solution