Q.Maximise and Minimise $Z = 3x - 4y$ subject to $x - 2y \le 0$, $-3x + y \le 4$, $x - y \le 6$, $x, y \ge 0$.

Graphical method. Optimise $Z=3x-4y$ over

Rewrite as $y\ge \frac{x}{2},\ y\le 3x+4,\ y\ge x-6$. The feasible region is unbounded with corner points $(0,0),\ (0,4),\ (12,6)$.

Maximum: the half-plane $3x-4y>12$ misses the region, so $Z_{\max}=12$ at $(12,6)$. …

Set up the region

As boundaries: $y\ge \dfrac{x}{2}$ (from $x-2y\le 0$), $y\le 3x+4$ (from $-3x+y\le 4$), and $y\ge x-6$ (from $x-y\le 6$).

Corner points

• $x=0$ with $-3x+y=4$: $(0,4)$.
• $x=0,\ y=0$: $(0,0)$.
• $x-2y=0 \cap x-y=6$: from $x=2y$, $2y-y=6\Rightarrow y=6,\ x=12$: $(12,6)$.

Each satisfies every constraint, so the vertices are $(0,0),\ (0,4),\ (12,6)$. The region opens upward (you can increase $y$ forever between $y=\frac{x}{2}$ and $y=3x+4$), so it is unbounded.

Evaluate $Z=3x-4y$

Maximum vs minimum on an unbounded region …

Method: Graphical (Corner-Point) Method for a Two-Variable LPP

Once a problem is written as "optimise $Z = ax + by$ subject to linear inequalities in $x$ and $y$," this is the standard CBSE technique for finding the optimum.

Steps

Step 1: Draw every constraint as a line.

Replace each inequality by an equation and plot the line from its axis intercepts. Include $x = 0$ and $y = 0$.

Step 2: Shade the correct half-plane.

Test the origin $(0,0)$ in each inequality: if it is satisfied, keep the origin's side; if not, take the other side. Non-negativity confines you to the first quadrant.

Step 3: Identify the feasible region.

It is the single region where all the shaded half-planes overlap.

Step 4: Find the corner points.

Each vertex is the intersection of two boundary lines, found by solving those two equations together. Keep a candidate only if it satisfies every other constraint -- an intersection that breaks a third inequality is not a corner of the region.

Step 5: Evaluate $Z$ at each corner.

By the Corner-Point Theorem the optimum of a linear $Z$ always occurs at a vertex, so compare $Z$ at the corners and pick the largest value (for a maximum) or the smallest (for a minimum).

Applying it here (asking for BOTH a maximum and a minimum) …

Common Mistakes

Mistake 1: Reporting the smallest corner value as the minimum on an unbounded region.

Why it's wrong: the least corner value is $-16$ at $(0,4)$, but along the open edge $y=3x+4$ the objective is $Z=-9x-16 \to -\infty$, so no minimum exists. Correct approach: run the open-half-plane check $3x-4y < (\text{least corner})$ -- since it meets the region, conclude the minimum does not exist.

Mistake 2: Sign slips when rewriting the inequalities for $y$. …