Consider the following L.P.P.
Minimize $z = 400x + 300y$, subject to $100x + 200y ≥ 12000, 300x + 400y ≥ 20000, 200x + 100y≥ 15000$ and $x, y ≥ 0$. Then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → its feasible region is unbounded and optimal point is (60, 30)
Given: Minimize $z=400x+300y$ subject to
$100x+200y\ge 12000,\quad 300x+400y\ge 20000,\quad 200x+100y\ge 15000,\quad x,y\ge 0$
Divide each inequality by 100 for convenience:
$x+2y\ge 120,\quad 3x+4y\ge 200,\quad 2x+y\ge 150$
Find intersection of the first and third equalities:
Solve $x+2y=120$ and $2x+y=150$ simultaneously ⇒ $(x,y)=(60,30)$.
Evaluate objective at candidate corner points of the feasible region:
At $(60,30)$: $z=400(60)+300(30)=24000+9000=33000$
At intersection with x-axis (from first inequality) $(120,0)$: $z=400(120)=48000$
At intersection with y-axis (from third inequality) $(0,150)$: $z=300(150)=45000$
Conclusion: The minimum value of $z$ is $33000$, attained at $(60,30)$.
