Maximise z = x + y subject to x + y ≤ 1 , − 3 x + y ≥ 3 and x ≥ 0 , y ≥ 0 , if possible?

Given constraints are x + y ≤ 1 , − 3 x + y ≥ 3 and x ≥ 0 , y ≥ 0 To maximise z = x + y Converting the given inequalities into equations x + y = 1 , − 3 x + y = 3 and x = 0 , y = 0 x + y = 1 - 3 x + y = 3 x = 0 y = 1 y = 3 y = 0 x = 1 x = - 1 Solving x + y = 1 , - 3 x + y = 3 we get x = - 0 . 5 , y = 1 . 5 The points are A 0 , 1 , B 1 , 0 , C - 1 , 0 , D 3 , 0 , E - 0 . 5 , 1 . 5 From the graph it is clear that there is no common region. So we cannot maximise the given constraints.

Solve the following linear programming problem graphically. Subject to the constraints : x + y ≤ 50 , 3 x + y ≤ 90 , x ≥ 0 , y ≥ 0 obtain the maximum and minimum values of Z = 5 x + 10 y .

Given objective function is Z = 5 x + 10 y for which we have to find maximum and values. The feasible region determined by x + y ≤ 50 , 3 x + y ≤ 90 , x ≥ 0 , y ≥ 0 is given by The corner points of feasible region are ( 0 , 0 ) , ( 0 , 50 ) , ( 20 , 30 ) , ( 30 , 0 ) . The value of Z at corner points are Corner Point Z = 5 x + 10 y ( 0 , 0 ) 0 Minimum ( 0 , 50 ) 500 Maximum ( 20 , 30 ) 400 ( 30 , 0 ) 150 The maximum value of Z is 500 and minimum is 0 .

Solve the following linear programming problem graphically. Subject to the constraints : x + y ≤ 50 , 3 x + y ≤ 90 , x ≥ 0 , y ≥ 0 obtain the maximum and minimum values of Z = 5 x + 10 y .

Given objective function is Z = 5 x + 10 y for which we have to find maximum and values. The feasible region determined by x + y ≤ 50 , 3 x + y ≤ 90 , x ≥ 0 , y ≥ 0 is given by The corner points of feasible region are ( 0 , 0 ) , ( 0 , 50 ) , ( 20 , 30 ) , ( 30 , 0 ) . The value of Z at corner points are Corner Point Z = 5 x + 10 y ( 0 , 0 ) 0 Minimum ( 0 , 50 ) 500 Maximum ( 20 , 30 ) 400 ( 30 , 0 ) 150 The maximum value of Z is 500 and minimum is 0 .

Maximise z = x + y subject to x + y ≤ 1 , − 3 x + y ≥ 3 and x ≥ 0 , y ≥ 0 , if possible?

Given constraints are x + y ≤ 1 , − 3 x + y ≥ 3 and x ≥ 0 , y ≥ 0 To maximise z = x + y Converting the given inequalities into equations x + y = 1 , − 3 x + y = 3 and x = 0 , y = 0 x + y = 1 - 3 x + y = 3 x = 0 y = 1 y = 3 y = 0 x = 1 x = - 1 Solving x + y = 1 , - 3 x + y = 3 we get x = - 0 . 5 , y = 1 . 5 The points are A 0 , 1 , B 1 , 0 , C - 1 , 0 , D 3 , 0 , E - 0 . 5 , 1 . 5 From the graph it is clear that there is no common region. So we cannot maximise the given constraints.