The solution of LPP
max.(z) = 5x + 3y subject to
\(2x+y \le 6\)
\(x+y \le 4\)
\(x \ge 0, y \ge 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → x = 2, y = 2, z = 16
The two boundary lines meet where
\(2x + y = 6, \quad x + y = 4\)
\(x = 2, \quad y = 2\)
The line 2x + y = 6 meets the x-axis at (3, 0), and x + y = 4 meets the y-axis at (0, 4). So the corner points of the feasible region are (0, 0), (3, 0), (2, 2) and (0, 4).
Values of z = 5x + 3y at the corners:
\(z(0, 0) = 0\)
\(z(3, 0) = 15\)
\(z(2, 2) = 10 + 6 = 16\)
\(z(0, 4) = 12\)
The maximum is z = 16 at x = 2, y = 2.