Optimise \(Z = 3x + 9y\) subject to the constraints:
\(x + 3y \le 60,\ x + y \ge 10,\ x \le y,\ x \ge 0,\ y \ge 0\), then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Maximum value of Z occurs at all the points on the line segment joining (15, 15) and (0, 20).
The feasible region is bounded, with corner points where the boundary lines meet:
\(x + y = 10,\ x = 0:\ (0, 10)\)
\(x + y = 10,\ x = y:\ (5, 5)\)
\(x + 3y = 60,\ x = y:\ (15, 15)\)
\(x + 3y = 60,\ x = 0:\ (0, 20)\)
Values of Z = 3x + 9y at the corners:
\(Z(0, 10) = 90\)
\(Z(5, 5) = 60\)
\(Z(15, 15) = 45 + 135 = 180\)
\(Z(0, 20) = 180\)
The maximum 180 is reached at two adjacent corners, so it is reached at every point of the edge joining them. Indeed
\(Z = 3x + 9y = 3(x + 3y)\)
and \(x + 3y = 60\) all along that edge, so Z = 180 at every point of the segment from (15, 15) to (0, 20), not just at its two ends.
NCERT (2026-27 edition):
Graphical Method of Solving Linear Programming Problems
