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+1 vote
2.7k views
in Linear Programming by (58.4k points)

Minimise and Maximise Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0,x,y ≥ 0.

1 Answer

+2 votes
by (53.2k points)
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Best answer

x + 2y = 120

x 60 0 20
y 30 60 50

0 + 0 < 120 True x + y = 60

x 30 20 40
y 30 40 20

0 + 0  0 false

x – 2y =0

x 20 0 40
y 10 0 20

40 – 20 > 0 

ABCD is the solution region 

At A (60, 30) Z = 5 x 60 + 10 x 30 = 600 

At B (40, 20) Z = 5 x 40+10 x 20 = 400 

At C (60,0) Z = 5 x 60 = 300 

At D (120,0) Z = 5 x 120 = 600 

Z is minimized at C (60,0) = 300 

Z is maximized at 2 points A (60,30), D (120,0) 

∴ maximum value is 600 at all 

points on the line joining A (60,30) and D (120,0)

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