Q.The feasible region of a linear programming problem is the bounded polygon with corner points $(0, 0)$, $(0, 8)$, $(4, 10)$, $(6, 8)$, $(6, 5)$ and $(5, 0)$. Let $Z = 3x - 4y$ be the objective function. The minimum of $Z$ occurs at
(A) $(0, 0)$
(B) $(0, 8)$
(C) $(5, 0)$
(D) $(4, 10)$

The Graphical Method for Linear Programming

When a linear programming problem has just two decision variables, $x$ and $y$, you can solve it by drawing a picture. This is the graphical method, and it is the technique the CBSE Class-12 course expects you to use.

The idea

Each constraint is a linear inequality such as $2x + 3y \le 100$. On the $xy$-plane its boundary is a straight line, and the inequality picks one side of that line (a half-plane). The points that satisfy all the constraints at once form a single region — the feasible region. Your job is to find the point inside this region that makes the objective function $Z = ax + by$ largest or smallest.

The step-by-step procedure

1. Draw each constraint line. Replace every inequality by an equation and plot the line, usually by finding where it meets the axes.
2. Shade the correct side. Test a simple point (often the origin $(0,0)$) in the inequality. If it holds, the origin's side is the wanted half-plane; if not, take the other side. Always include the non-negativity conditions $x \ge 0,\ y \ge 0$, which keep you in the first quadrant.
3. Identify the feasible region. It is the overlap of all the shaded half-planes — the region satisfying every constraint together.
4. Find the corner (vertex) points. These are the points where the boundary lines cross. Read them off the graph or solve the two relevant lines simultaneously.
5. Evaluate $Z$ at every corner and pick the largest value (for a maximum) or the smallest (for a minimum).

Bounded vs unbounded …