## Solve the following problem graphically: Minimise and Maximise z = 3 x + 9 y Subject to the constraints: x + 3 y ≤ 60 x + y ≥ 10 x ≤ y x ≥ 0 , y ≥ 0 .

Let z = 3 x + 9 y . . . . ( 1 ) converting inequalities to equalities x + 3 y = 60 x 0 60 y 20 0 points are ( 0 , 20 ) , ( 60 , 0 ) x + y = 10 x 0 10 y 10 0 points are ( 0 , 10 ) , ( 10 , 0 ) x − y = 0 x 0 10 20 y 0 10 20 points are ( 0 , 0 ) , ( 10 , 10 ) , ( 20 , 20 ) plot the graph for the set of points the graph shows the bounded feasible region. a b c d , with corner points a = ( 10 , 0 ) , b = ( 5 , 5 ) , c = ( 15 , 15 ) and d = ( 0 , 20 ) to find maximum and minimum corner point z = 3 x + 9 y a = ( 0 , 10 ) 90 b = ( 5 , 5 ) 60 c = ( 15 , 15 ) 180 d = ( 0 , 20 ) 180 from the graph maximum value of x occurs at two corner points c ( 15 , 5 ) and d ( 0 , 20 ) with value 180 and minimum occurs at point b ( 5 , 5 ) with value 60 ..

Find Minimize Z=3x+9y subject to

x+3y≤60 , x≤y and x,y≥0

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## Solve the following problem graphically: Minimize and Maximize Z = 3 x + 9 y Subject to the constraints: x + 3 y ≤ 60 x + y ≥ 10 x ≤ y x ≥ 0 , y ≥ 0

Maximise and minimise : z = 3 x + 9 y subject to x + 3 y ≤ 60 , x + y ≥ 10 , x ≤ y x ≥ 0 , y ≥ 0 x + 3 y = 60 x 0 60 y 20 0 x + y = 10 x 0 10 y 10 0 x = y x 0 20 y 0 20 corner points value of z = 3 x + 9 y ( 0 , 10 ) 90 ( 5 , 5 ) 60 ( 15 , 15 ) 180 ( 0 , 20 ) 180 ∴ z = 60 is minimum at ( 5 , 5 ) also, z is maximum at two points ( 15 , 15 ) & ( 0 , 20 ) ∴ z = 180 is maximum at all points joining ( 15 , 15 ) & ( 0 , 20 ) ..

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