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The maximum value of a LPP z=3x+4yz = 3x + 4y subject to the constraints: x+y≤6x + y \leq 6, x≥0x \geq 0, y≥0y \geq 0 is:

Solution

✅ Correct Option: 1

The constraints are:

x+y≤6x + y \leq 6

x≥0x \geq 0

y≥0y \geq 0

These create a triangular feasible region.


The corner points of the feasible region are found where the constraint lines intersect:

At x=0x = 0 and y=0y = 0: (0,0)(0, 0)

At y=0y = 0 and x+y=6x + y = 6:

x+0=6x + 0 = 6

x=6x = 6

Corner point: (6,0)(6, 0)

At x=0x = 0 and x+y=6x + y = 6:

0+y=60 + y = 6

y=6y = 6

Corner point: (0,6)(0, 6)


Evaluating z=3x+4yz = 3x + 4y at each corner point:

At (0,0)(0, 0):

z=3(0)+4(0)z = 3(0) + 4(0)

z=0z = 0

At (6,0)(6, 0):

z=3(6)+4(0)z = 3(6) + 4(0)

z=18z = 18

At (0,6)(0, 6):

z=3(0)+4(6)z = 3(0) + 4(6)

z=24z = 24


The maximum value of zz is 2424, occurring at the point (0,6)(0, 6).

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