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Miscellaneous Exercise 7 · Q55

Q.Solution of L.P.P. to minimize z=2x+3yz = 2x + 3y s.t. x≥0, y≥0, 1≤x+2y≤10x \ge 0,\ y \ge 0,\ 1 \le x + 2y \le 10 is _______. A) x=0, y=12x = 0,\ y = \dfrac{1}{2}
B) x=12, y=0x = \dfrac{1}{2},\ y = 0
C) x=1, y=2x = 1,\ y = 2
D) x=12, y=12x = \dfrac{1}{2},\ y = \dfrac{1}{2}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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The constraint 1≤x+2y≤101\le x+2y\le10 splits into x+2y≥1x+2y\ge1 and x+2y≤10x+2y\le10; together with x≥0,y≥0x\ge0,y\ge0 the feasible vertices are (1,0)(1,0), (10,0)(10,0), (0,5)(0,5), and (0,12)\left(0,\tfrac12\right). Evaluating z=2x+3yz=2x+3y: at (1,0)(1,0), z=2z=2; at (10,0)(10,0), z=20z=20; at (0,5)(0,5), z=15z=15; at (0,12)\left(0,\tfrac12\right), …

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