linear programming by graphical method
Maximise z = 100x + 150y
subject to
3x + 5y <= 3900
x + 3y <= 2100
2x + 2y <= 2200
x >= 0 and y >= 0
3x + 5y = 3900 passes throungh (1300,0 ) and (0,780)
x + 3y = 2100 passes throungh (2100,0 ) and (0,700)
2x + 2y = 2200 passes throungh (1100,0 ) and (0,1100)
the non negative conditions imply the first quadrant
Mapping the given inequalities
The shaded region OABCD is the feasible region
The vertices are O(0,0)
A(1100,0 )
B(800,300)
C(300,600)
D(0,700)
z = 100x + 150y
at O(0,0) , z = 0
at A(1100,0 ), z = 110000
at B(800,300), z = 125000
at C(300,600) , z = 120000
at D(0,700) , z = 105000
therefore maximum value of z is 125000 when x=800 and y=300
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