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Show that the minimum of Z occurs at more than two points.
Minimise and maximise $$Z = x + 2y$$subject to $$x + 2 y \geq 100,2 x - y \leq 0,2 x + y \leq 200$$ $$x , y \geq 0$$
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## QuestionMathsClass 12

Show that the minimum of Z occurs at more than two points.
Minimise and maximise $$Z = x + 2y$$subject to $$x + 2 y \geq 100,2 x - y \leq 0,2 x + y \leq 200$$ $$x , y \geq 0$$

see the analysis below
4.6
4.6

## Solution

We have linear constraints as
$$x + 2 y \geq 100$$
$$2 x - y \leq 0$$
$$2x+y\leq 200$$
$$x \geq 0 , y \geq 0$$
and objective function is min or max $$(Z) = x + 2y$$
Now, reducing the above inequations into equations and finding their point of intersections i.e.,
$$x + 2y = 100$$ ...(i)
$$2x - y = 0$$ ...(ii)
$$2x + y = 200$$ ...(iii)
$$x = 0, y = 0$$..(iv)

Now for feasible region, using origin testing method for each constraint
For $$x + 2 y \geq 100$$, let $$x = 0, y = 0$$
$$\Rightarrow 0 \geq 100$$ i.e., false $$\Rightarrow$$ the shaded region will be away from the origin
For $$2 x + y \leq 200$$, let $$x = 0, y = 0$$$$\Rightarrow 0 \leq 200$$ i.e., true $$\Rightarrow$$ The shaded region will be toward the origin.
For $$2 x- y \leq 0$$, let $$x = 0, y = 0$$$$\Rightarrow 0 \leq 0$$ i.e., true $$\Rightarrow$$ The shaded region will be toward the origin.
Also, non-negative restrictions $$x \geq 0 , y \geq 0$$ indicates that the feasible region will be exist in first quadrant.
Now, corner points are $$A(0, 50), B(20, 40), C(50, 100)\ and\ D(0, 200)$$

For optimal solution substituting the value of all comer points in $$Z=x+2y$$

Hence, minimum $$(Z) = 100$$ at all points on the line segment joining the points $$A(0, 50)$$and $$B(20, 40)$$ : Maximum $$(Z) = 400$$at $$(0, 200)$$