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Question

Minimize and miximize Z = x + 2y subject to constraints are x + 2y 100, 2x - y 0, 2x + y 200 and x, y  0.


Solution

Our problem is to minimize and miximize 

Z = x + 2y ...(i)

Subject to constraints are x + 2y 100 .....(ii)

2x - y  0 ...(iii)

2x + y  200   .....(iv)

0, y  0 .....(v)

Firstly, draw the graph of the line x + 2y = 100

x0100y500

Putting (0, 0) in the inequality x + 2y 100, we have

0+2×0100

0100     (which is false)

So, the half plane is away from the origin.

Secondly, draw the graph of line 2x - y = 0

x010y020

Putting (5, 0) in the inequality 2x - y 0, we have 2×500100    (which is false)

So, the half plane is towards Y-axis.

Thirdly, draw the graph of line 2x + y = 200

x0100y2000 

Putting (0, 0) in the inequality 2x + y 200, 

we have 2×0+02000200 (which is true)

So, the half plane is towards the origin, Since, x, y geq 0

So, the feasible region lies in the first quadrant.

On solving equations 2x - y = 0 and x + 2y = 100, we get B(20, 40) and on solving equations 2x - y = 0 and x + 2y = 200, we get C(50, 100)

Feasible region is ABCDA,

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 points are as follows:

Corner pointZ=x+2yA(0, 50)100MinimumB(20, 40)100MinimumC(50,100)250D(0,200)400Minimum

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).

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