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The maximum value of the objective function Z=2x+yZ = 2x + y of an LPP, subject to the constraints x≤6,y≤2,x−y≤0x \leq 6, y \leq 2, x - y \leq 0, x≥0,y≥0x \geq 0, y \geq 0 is

Solution

✅ Correct Option: 2

The constraints are:

x≤6x \leq 6

y≤2y \leq 2

x−y≤0x - y \leq 0 (which means x≤yx \leq y)

x≥0x \geq 0

y≥0y \geq 0


In Linear Programming, the maximum or minimum value occurs at a corner point of the feasible region.

The corner points are found by checking where the boundary lines intersect:

Origin: (0,0)(0, 0)

Where x=0x = 0 meets y=2y = 2: (0,2)(0, 2)

Where x=yx = y meets y=2y = 2:

If y=2y = 2 and x=yx = y, then x=2x = 2

Point: (2,2)(2, 2)

Note: The constraint x≤6x \leq 6 is not active since x≤yx \leq y and y≤2y \leq 2 means xx cannot exceed 22.


Evaluating Z=2x+yZ = 2x + y at each corner point:

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

Z=2(0)+0=0Z = 2(0) + 0 = 0

At (0,2)(0, 2):

Z=2(0)+2=2Z = 2(0) + 2 = 2

At (2,2)(2, 2):

Z=2(2)+2Z = 2(2) + 2

Z=4+2Z = 4 + 2

Z=6Z = 6


Comparing all values: 00, 22, and 66

The maximum value is 66 at point (2,2)(2, 2).

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