1. **Problem Statement:** Maximise the objective function $$Z = 3x + 2y + 1$$ subject to the constraints $$x \geq 0$$, $$y \geq 0$$, and $$3x + 4y \leq 12$$. We need to identify the missing constraint from the options given. 2. **Understanding Constraints:** The feasible region is bounded by the axes $$x=0$$, $$y=0$$, the line $$3x + 4y = 12$$, and one more constraint that forms a triangular feasible region. 3. **Given Options:** - (A) $$x + 2y \leq 2$$ - (B) $$2x + y \geq 2$$ - (C) $$2x + y \leq 2$$ - (D) $$x + 2y \geq 2$$ 4. **Analyzing the Graph Description:** The additional diagonal line intersects the feasible region and creates a triangular shaded area. This means the missing constraint must be a line that cuts through the region and restricts it further. 5. **Testing Each Option:** - (A) $$x + 2y \leq 2$$: This line passes through points (2,0) and (0,1). It lies inside the first quadrant and can form a boundary. - (B) $$2x + y \geq 2$$: This inequality represents the region above the line $$2x + y = 2$$, which would expand the feasible region, not restrict it. - (C) $$2x + y \leq 2$$: This line passes through points (1,0) and (0,2), restricting the region inside the first quadrant. - (D) $$x + 2y \geq 2$$: This represents the region above the line $$x + 2y = 2$$, which would not restrict the feasible region to a triangle. 6. **Comparing with the Graph:** The diagonal line forming the triangular region is likely to be $$2x + y \leq 2$$ (Option C), as it fits the description of intersecting the feasible region and creating a bounded triangle with the axes and $$3x + 4y = 12$$. **Final answer:** The missing constraint is (C) $$2x + y \leq 2$$.