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In a linear programming problem(LPP), the maximum value of the objective function Z=2x+5yZ = 2x + 5y subjected to the constraints:

2x+3y≤62x + 3y ≤ 6

2x+y≤42x + y ≤ 4

x,y≥0x, y ≥ 0

is

Solution

✅ Correct Option: 2

The constraints create a bounded feasible region. The maximum value of the objective function occurs at one of the corner points of this region.


The corner points are found by examining intersections of the constraint boundaries.

At the origin:

Point: (0,0)(0, 0)


On the xx-axis, set y=0y = 0:

From 2x+3y≤62x + 3y \leq 6: 2x≤62x \leq 6, so x≤3x \leq 3

From 2x+y≤42x + y \leq 4: 2x≤42x \leq 4, so x≤2x \leq 2

The stricter constraint gives x=2x = 2

Point: (2,0)(2, 0)


On the yy-axis, set x=0x = 0:

From 2x+3y≤62x + 3y \leq 6: 3y≤63y \leq 6, so y≤2y \leq 2

From 2x+y≤42x + y \leq 4: y≤4y \leq 4

The stricter constraint gives y=2y = 2

Point: (0,2)(0, 2)


At the intersection of 2x+3y=62x + 3y = 6 and 2x+y=42x + y = 4:

2x+3y=62x + 3y = 6

2x+y=42x + y = 4

Subtracting the second from the first:

2y=22y = 2

y=1y = 1

Substituting into 2x+y=42x + y = 4:

2x+1=42x + 1 = 4

2x=32x = 3

x=1.5x = 1.5

Point: (1.5,1)(1.5, 1)


Evaluating Z=2x+5yZ = 2x + 5y at each corner point:

At (0,0)(0, 0): Z=2(0)+5(0)=0Z = 2(0) + 5(0) = 0

At (2,0)(2, 0): Z=2(2)+5(0)=4Z = 2(2) + 5(0) = 4

At (0,2)(0, 2): Z=2(0)+5(2)=10Z = 2(0) + 5(2) = 10

At (1.5,1)(1.5, 1): Z=2(1.5)+5(1)=3+5=8Z = 2(1.5) + 5(1) = 3 + 5 = 8


The maximum value of ZZ is 1010, occurring at the point (0,2)(0, 2).

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