The maximum value of Z = 5 x + 4 y , Subject to y ≤ 2 x , x ≤ 2 y , x + y ≤ 3 , x ≥ 0 , y ≥ 0 is ..

Step-by-step solution

We have, z = 5 x + 4 y Subject to contraints y ≤ 2 x , x ≤ 2 y , x + y ≤ 3 , x ≥ 0 , y ≥ 0 On taking given constraints as equations, we get the following3 graph. Intersecting point of line y = 2 x and x + y = 3 is A 1 , 2 and intersecting point of line y = 2 y and x + y = 3 is B 2 , 1 Here OABO is the required feasible region whose corner points are O 0 , 0 , A 1 , 2 and B 2 , 1 Corner points z = 5 x + 4 y O , 0,0 5 × 0 + 4 × 0 = 0 A 1,2 5 × 1 + 4 × 2 = 13 B 2,1 5 × 2 + 4 × 1 = 14 (maximum)