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Question

Minimise and Maximise Z=x+2y
subject to x+2y100,2xy0,2x+y200;x,y0

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Solution

The feasible region determined by the constraints x+2y100,2xy0,2x+y200;x0,y0 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 pointZ=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).
457560_422953_ans.PNG

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