1.

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)

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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