The maximum value of the objective function \(Z = 10x + 15y\) of an L.P.P, subjected to the constraints
\(2x + 4y \le 8\),
\(3x + y \le 6\),
\(-x - y \ge -4\),
\(x \ge 0,\ y \ge 0\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 34
The constraints simplify to
\(x + 2y \le 4\)
\(3x + y \le 6\)
\(x + y \le 4\)
with \(x, y \ge 0\). For \(x, y \ge 0\), \(x + y \le x + 2y \le 4\), so the third constraint is redundant. The lines \(x + 2y = 4\) and \(3x + y = 6\) meet where
\(x + 2(6 - 3x) = 4\)
\(5x = 8,\quad x = 1.6,\quad y = 1.2\)
By the corner point method, with corners (0, 0), (2, 0), (1.6, 1.2) and (0, 2):
\(Z(0, 0) = 0\)
\(Z(2, 0) = 20\)
\(Z(1.6, 1.2) = 16 + 18 = 34\)
\(Z(0, 2) = 30\)
The maximum value is 34, at (1.6, 1.2).