The number of feasible solutions of the linear programming problem as
Maximum \( z=15x+30y \) subject to the constraints
\( 3x+y\le 12 \)
\( x+2y\le 10 \)
\( x\ge 0,y\ge 0 \)
is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Infinite
A feasible solution is any point that satisfies all the constraints, not only a corner point.
The constraints are satisfied by (0, 0), so the feasible region is not empty. Its corner points are
\( \displaystyle (0,0),\ (4,0),\ \left(\frac{14}{5},\frac{18}{5}\right),\ (0,5) \)
where \( \left(\frac{14}{5},\frac{18}{5}\right) \) comes from solving \( 3x+y=12 \) and \( x+2y=10 \).
The region is the whole quadrilateral with these corners, including its interior, so it contains infinitely many points, for example \( (t,t) \) for every t in [0, 1].
So the number of feasible solutions is infinite.
Tip: Do not confuse feasible solutions (every point of the region) with optimal solutions. Here even the optimum is not unique: z = 15(x + 2y), so z = 150 at every point of the edge on x + 2y = 10.