The L.P.P. , minimize z=30 x+20 y, x+y 8, x+2 y 4, 6 x+4 y 12 , x 0, y 0 has

Step-by-step solution

Linear Programming Problem: Minimize z = 30x + 20y subject to the constraints: x + y 8 x + 2y 4 3x + 2y 6 x 0 , y 0 The feasible region lies in the first quadrant, bounded by the lines x + y = 8 , x + 2y = 4 , and 3x + 2y = 6 . Intersections of these constraints yield the vertices: From x = 0 and 3x + 2y = 6 , y = 3 : (0, 3) . From x = 0 and x + y = 8 , y = 8 : (0, 8) . From y = 0 and x + 2y = 4 , x = 4 : (4, 0) . From y = 0 and x + y = 8 , x = 8 : (8, 0) . From x + 2y = 4 and 3x + 2y = 6 , solving gives x = 1 , y