Solve the following Linear Programming Problem Graphically. Maximize $Z = 5x + 3y$ Subject to constraints: $3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0$, and $y ≥ 0$.

12.37

The correct answer is Option (3) → 12.37

By plotting the given linear inequalities, we can see that the inequality 3x + 5y ≤ 15 meets the co-ordinates axes at points (5, 0) and A(0, 3) respectively.

Also the inequality 5x + 2y ≤ 10 meets the co-ordinates axes at points C(2,0) and (0,5) respectively.

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

The coordinates of the vertices (corner point) of the shaded bounded feasible region are O (0, 0), A (0, 3), B (20/19, 45/19) and C (2, 0).

These points have been obtained by solving the equations of the corresponding intersecting lines, simultaneously. The value of the objective function as these points are given in the following table:

Clearly, Z is the maximum at P (20/19, 45/19)

Hence, x = 20/19, y = 45/19 is the optimal solution of the given LPP.

The optimal maximum value of Z is 12.37 when x = 20/19 and y = 45/19