1. The problem states that the feasible region of a linear programming problem is bounded and the objective function attains its minimum value at more than one point. 2. One of these points is given as (5,0). 3. Since the objective function attains the minimum at multiple points, the minimum value lies along a line segment connecting these points. 4. The point (5,0) lies on the x-axis, so the other point must also lie on the same boundary line where the objective function is constant. 5. From the graph description, the feasible region is a triangle bounded by three lines, one of which is horizontal just above the x-axis from around (4,0) to (10,0). 6. Since (5,0) is on this horizontal boundary, the objective function is constant along this line segment. 7. Therefore, the other point where the minimum is attained must also lie on this horizontal line segment. 8. Among the options, (2,9), (6,6), (4,7), and (0,0), only (6,6) and (4,7) are not on the x-axis, and (0,0) and (2,9) are not on the horizontal boundary near (5,0). 9. The horizontal boundary near (5,0) extends roughly from (4,0) to (10,0), so the other point must be on this line segment. 10. The only point on this line segment from the options is (0,0) is not on the segment, (2,9) and (4,7) are above the x-axis, (6,6) is above the x-axis. 11. Since the line segment is horizontal at y=0, the other point must have y=0 and be between x=4 and x=10. 12. The only candidate is (4,0), but it is not given as an option. 13. Since (5,0) is on the boundary and the minimum is attained at more than one point, the other point must be on the same boundary line. 14. The closest option that fits is (0,0) which lies on the boundary line rising steeply from (0,0) to (1,16), but it is not on the same horizontal boundary. 15. Given the options, the best choice is (0,0) because it lies on the boundary and could be the other point where the minimum is attained. Final answer: (D) (0,0)