To solve this problem, we need to first plot the given constraints and identify the feasible region:
Constraint 1: x + y ≥ 10. This is a straight line with slope -1. The feasible region is above this line.
Constraint 2: 2x + 2y ≤ 25. Simplifying this inequality, we get x + y ≤ 12.5, which is a straight line with slope -1. The feasible region is below this line.
Constraint 3: x ≥ 0. This constraint restricts the feasible region to the right of the y-axis.
Constraint 4: y ≥ 0. This constraint restricts the feasible region to above the x-axis.
Thus, the feasible region is the intersection of the regions defined by these four constraints.
The graph that correctly represents this feasible region is shown in Option (3).
List-I | List-II | ||
A | If the corner points of the feasible region For an LPP are (0, 4), (5, 0), (7, 9), then the minimum value of the objective function Z =5x+y is. | I | 27 |
B | If the corner points of the feasible region for an LPP are (0, 0), (0, 2), (3, 4), (5, 3). then the maximum value of the objective function Z=3x+4y | II | 60 |
C | The comer points of the feasible region for an LPP are (0, 2), (1, 2), (4,3), (7, 0). The objective function is Z = x+5y. Then (Max Z+Min Z) is | III | 25 |
D | If the corner points of the feasible region for an LPP are (0, 4), (3, 0), (3, 2), (6,9) The objective function is Z=2x+6y. Then (Max Z-Min Z) | IV | 26 |