Practicing Success
Consider the LPP, Max z = 2x + 3y, subject to the conditions, x + y ≤ 2, x ≤ 2 x ≥ 0 ; y ≥ 0, then maximum value of the objective function is : |
8 6 4 0 |
6 |
$x \geq 0, y \geq 0$ $x \leq 2$ $x+y \leq 2$ Function to be maximised Z = 2x + 3y x ≥ 0, y ≥ 0 solution is in first quadrant plotting for x + y = 2
now checking for (0, 0) in x + y ≤ 2 0 ≤ 2 solution lies in part containing (0, 0) below line corner points obtained → A (0, 0) B (0, 2) C (2, 0) Z (x, y) = 2x + 3y Z(0, 0) = 2(0) + 3(0) = 0 Z (0, 2) = 2(0) + 3(2) = 0 + 6 = 6 Z (2, 0) = 2(2) + 3(0) = 4 + 0 = 4 for (B (0, 2)) Z is maximum Zmax = 6 |