The point/points which provides the optimal solution of the linear programming problem maximum \( z=3x+y \), \( x+3y\le 12,3x+y\le 12,x,y\ge 0, \) is/are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Points on line segment joining (4, 0) and (3, 3).
The lines x + 3y = 12 and 3x + y = 12 meet at (3, 3). The feasible region has corner points (0, 0), (4, 0), (3, 3) and (0, 4). Values of z = 3x + y:
\( \displaystyle (0,\,0):\ z=0 \)
\( \displaystyle (4,\,0):\ z=12 \)
\( \displaystyle (3,\,3):\ z=12 \)
\( \displaystyle (0,\,4):\ z=4 \)
The maximum 12 occurs at two adjacent corners, (4, 0) and (3, 3). The objective function 3x + y is the left side of the constraint 3x + y ≤ 12, so z = 12 at every point of the edge joining them.
Hence every point on the line segment joining (4, 0) and (3, 3) is an optimal solution.