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) x ≥ 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+2×0≥100 ⇒0≥100     (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 2×5−0≤0⇒10≤0    (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+0≤200⇒0≤200 (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)100→MinimumB(20, 40)100→MinimumC(50,100)250D(0,200)400→Minimum 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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