Choose the correct alternative : Solution of LPP to minimize z = 2x + 3y st. x $\geq$ 0, y $\geq$ 0, 1$\leq$ x + 2y $\leq$ 10 is

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Z = 2x + 3y The given inequalities are 1 $\leq$ x + 2y $\leq$ 10 i.e. x + 2y $\geq$ 1 and x + 2y $\leq$ 10 consider lines L1 and L2 where L1 : x + 2y = 1, L2 : x + 2y = 10. For line L1 plot A(0, $\frac{1}{2}$), B(1, 0) For line L2 plot P (0, 5), Q (10, 0). The coordinates of origin O (0, 0) do not satisfy x + 2y $\geq$ 1. Required region lies on non – origin side of L1. The coordinates of origin O(0, 0) satisfies the inequalities x + 2y $\leq$ 10. Required region lies on the origin side of L2. Lines L1 and L2 are parallel. ABQPA is the required feasible region At "A"(0, $\frac{1}{2}$), "Z" = 0+ 3($\frac{1}{2}$) = 1.5 At B (1, 0), Z = 2 (1) + 0 = 2 At P (0, 5), Z = 0 + 3(5) = 15 At Q (10, 0), Z = 2 (10) + 0 = 20 The maximum value of Z is 1.5 and it occurs at "A"(0, $\frac{1}{2}$) i.e. x = 0, y = (1)/(2)

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