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The feasible solution for a LPP is shown in Fig. Let Z = 3x – 4y be the objective function. Maximum of Z occurs atA. (5, 0)B. (6, 5)C. (6, 8)D. (4, 10) |
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Answer» Value of Z = 3x – 4y, at corner points are – At (0, 0) = 0 At (0, 8) = -32 At (4, 10) = -28 At (6, 8) = -14 At (6, 5) = -2 At (5, 0) = 15 So, clearly maximum value is at (5, 0) Ans is A ( 5 , 0 )X Y 3X-4Y 0 8 -32 4 10 -28 6 8 -14 6 5 -2 0 0 0 5 0 15 |
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