Solve the following Linear Programming Problem Graphically. Maximize $Z = 2x + 4y$ Subject to constraints: $x + 2y ≤ 5, x + y ≤ 4, x ≥ 0$ and $y ≥ 0$.

10

The correct answer is Option (2) → 10

By plotting the given linear inequalities, we can see that the inequality $x + 2y ≤ 5$ meets the co-ordinates axes at the point (5, 0) and A(0, 2.5) respectively.

Similarly, The inequality $x + y ≤ 4$ meets the co-ordinates axes at the point C (4, 0) and (0, 4)

As shown in the graph above, the shaded bounded region OABCO represents the common region of the above inequation. This region is the feasible region of the given LPP.

The coordinates of the vertices (corner point) of the shaded feasible region are O (0, 0), A (0, 2.5), B (3, 1) and C (4, 0).

The value of the objective function as these corner points are given in the following table:

Clearly, Z has maximized at two corner points A (0, 2.5) and B (3, 1).

Hence, any point on the line segment joining points A and B will give the maximum value $Z = 10$ of the objective function.

The optimal maximised value of Z is 10 when $x = 0$ and $y = 2.5$ or when $x = 3$ and $y = 1$.