The optimal value of linear programming problem maximum Z = 3x + 4y, subject to,
\(x + 3y \le 12\)
\(x + y \ge 8\)
\(x, y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 36
The boundary lines x + 3y = 12 and x + y = 8 meet where
\((x + 3y) - (x + y) = 12 - 8\)
\(2y = 4,\quad y = 2,\quad x = 6\)
On the x-axis (y = 0) the constraints give \(x \ge 8\) and \(x \le 12\). The points must lie on or below x + 3y = 12 and on or above x + y = 8, so the feasible region is the triangle with corners (8, 0), (12, 0) and (6, 2).
Values of Z = 3x + 4y at the corner points:
\(Z(8, 0) = 24\)
\(Z(12, 0) = 36\)
\(Z(6, 2) = 18 + 8 = 26\)
The maximum value is 36, at (12, 0).