Minimise and Maximise Z=x+2y subject to x+2y≥100,2x−y≤0,2x+y≤200;x,y≥0
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Solution
The feasible region determined by the constraints x+2y≥100,2x−y≤0,2x+y≤200;x≥0,y≥0 is as shown. The corner points of the feasible region are A(0,50),B(20,40),C(50,100) and D(0,200) The values of Z at these corner points are as follows.
Corner point
Z=x+2y
A(0,50)
100
→ Minimum
B(20,40)
100
→ Minimum
C(50,100)
250
D(0,200)
400
→ Maximum
The
maximum value of Z is 400 at (0,200) and the minimum value
of Z is 100 at all the points on the line segment joining the
points (0,50) and (20,40).