Graphical solution: feasible region

A linear inequality $ax + by \le c$ divides the $xy$ -plane into two half-planes. The boundary $ax + by = c$ is a line. The feasible region of an LPP is the intersection of all such half-planes (one per constraint), together with the first quadrant from the non-negativity constraints.

Graphing a linear inequality

Draw the boundary line $ax + by = c$ , a straight line through two convenient points (often the intercepts $(c/a, 0)$ and $(0, c/b)$ ).
Pick a test point not on the line , usually $(0, 0)$ if it's not on the line.
Substitute into the inequality. If it's satisfied, the test point is in the half-plane represented by the inequality. Shade that side.
Combine all such half-planes by overlapping shaded regions; the overlap is the feasible region.

For $\le$ inequalities, the boundary line is included; draw it solid. For $<$ , draw dashed. (In LP we mostly use $\le, \ge, =$ .)

Standard layout

A typical LPP in Class XII has:

Two decision variables $x, y$ (both $\ge 0$ ),
A few inequality constraints,
Sometimes equality constraints (which give a line, not a half-plane).

So the feasible region lives in the first quadrant. Sketch the axes; mark the intercepts of each constraint line; identify the side that satisfies each inequality; the overlap is the feasible region.

Bounded vs unbounded

A bounded feasible region is enclosed , a polygon. The corner-point method always yields a maximum and minimum for the linear objective.

An unbounded feasible region extends to infinity. The maximum (or minimum) may not exist; you must check whether the objective is bounded above (or below) on the region. For a maximisation problem with constraints $\ge$ , the region is often unbounded toward the upper-right; the maximum may be infinite.

Corner points

A corner point (vertex) of the feasible region is where two constraint lines meet (and the resulting point satisfies all other constraints). To find corners systematically:

Take each pair of constraint equations.
Solve simultaneously for the intersection point.
Check if the point satisfies all other inequalities.
Keep it if yes; discard otherwise.

Corner points always include those on the axes (intersections with $x = 0$ or $y = 0$ ) when those intersections are feasible.

Worked examples

Example 1. Find the feasible region: $x \ge 0$ , $y \ge 0$ , $x + y \le 4$ , $2x + y \le 6$ .

The boundary lines: $x + y = 4$ (intercepts $(4, 0), (0, 4)$ ); $2x + y = 6$ (intercepts $(3, 0), (0, 6)$ ). Both inequalities are $\le$ , with $(0, 0)$ on the correct side. Together with the first quadrant, the region is the polygon with corners at $(0, 0)$ , $(3, 0)$ , intersection of $x + y = 4$ and $2x + y = 6$ , $(0, 4)$ .

Intersect $x + y = 4, 2x + y = 6$ : subtract to get $x = 2, y = 2$ . So corners: $(0, 0), (3, 0), (2, 2), (0, 4)$ .

Example 2. Feasible region of $x + y \ge 2$ , $x \le 3$ , $y \le 3$ , $x, y \ge 0$ .

$x + y = 2$ (intercepts $(2, 0), (0, 2)$ ), with the region above (test $(0, 0)$ : $0 < 2$ , fails , so region excludes origin). $x \le 3$ and $y \le 3$ truncate.

Corners: where $x = 0, y = 3$ meet (i.e. $(0, 3)$ ); $(3, 3)$ (corner of the box); $(3, 0)$ ; $(0, 2)$ ( $x = 0, x + y = 2$ ); $(2, 0)$ ( $y = 0, x + y = 2$ ). So five corners: $(0, 2), (0, 3), (3, 3), (3, 0), (2, 0)$ .

Example 3. Feasible region of $x + y \ge 5$ , $2x + 3y \ge 12$ , $x, y \ge 0$ .

Both lines slope downward and the region is above both. Boundaries: $x + y = 5$ at $(5, 0), (0, 5)$ ; $2x + 3y = 12$ at $(6, 0), (0, 4)$ .

The region is unbounded , extends infinitely up-and-right. Corner points: where the two lines meet (solve: subtract twice the first from the second: $y = 2$ , so $x = 3$ ). On axes: $(6, 0)$ ( $y = 0$ on second line; check $6 + 0 \ge 5$ , yes) and $(0, 5)$ ( $x = 0$ on first; check $0 + 15 \ge 12$ , yes). So corners: $(6, 0), (3, 2), (0, 5)$ .

Example 4. Feasible region of $x + y \le 8$ , $x \ge 2$ , $y \ge 1$ , $x, y \ge 0$ .

Bounded polygon. Corners: at $x = 2, y = 1$ : $(2, 1)$ . At $x = 2, x + y = 8$ : $(2, 6)$ . At $y = 1, x + y = 8$ : $(7, 1)$ . Three corners , it's a triangle.

Example 5. Empty feasible region: $x + y \le 1$ , $x + y \ge 5$ . The first says $x + y$ small, the second says big , no overlap. Infeasible.

Example 6. Single point: $x + y = 4$ , $x - y = 0$ , $x, y \ge 0$ . Solve: $x = y = 2$ . Only one feasible point.

Try it yourself

For each, find the feasible region and its corners.

$x + y \le 4$ , $x \le 3$ , $y \le 3$ , $x, y \ge 0$ .
$2x + y \le 10$ , $x + 3y \le 12$ , $x, y \ge 0$ .
$x + y \ge 3$ , $x - y \ge -1$ , $x, y \ge 0$ .
$3x + 2y \ge 12$ , $x + 4y \ge 8$ , $x, y \ge 0$ .
$x + 2y \le 8$ , $3x + 2y \le 12$ , $x \ge 0$ , $y \ge 0$ .
$x \le 4$ , $y \le 6$ , $x + 2y \le 10$ , $x, y \ge 0$ .
$5x + y \ge 10$ , $x + y \ge 6$ , $x + 4y \ge 12$ , $x, y \ge 0$ .
$x + y \ge 8$ , $x - y \le 4$ , $x, y \ge 0$ .
$2x + y \ge 4$ , $x + y \le 5$ , $x \le 4$ , $x, y \ge 0$ .
$3x + 4y \le 60$ , $x + 3y \le 30$ , $x, y \ge 0$ .
$5x + 4y \le 20$ , $x + y \ge 1$ , $y \ge 2$ , $x \ge 0$ .
$x + 2y \le 10$ , $3x + y \le 15$ , $x + y \ge 2$ , $x, y \ge 0$ .
$4x + 3y \le 240$ , $x + 2y \le 80$ , $x, y \ge 0$ .
$2x + y \ge 6$ , $x - y \ge -1$ , $x + 2y \le 8$ , $x, y \ge 0$ .

Pitfalls and tricks

Always test which side of a boundary line satisfies the inequality. Don't guess.
Shade carefully. When constraints overlap, the feasible region is the intersection , keep only the multiply-shaded part.
Intersection points are found by solving pairs of constraint equations. Check that each candidate corner satisfies all other constraints.
Watch for unboundedness. An unbounded region may have no finite maximum (for a maximisation problem).
Sketch on graph paper or carefully on grid paper , accurate intercepts and intersections matter.