Q.Maximise $Z = x + y$ subject to $x + 4y \le 8$, $2x + 3y \le 12$, $3x + y \le 9$, $x \ge 0$, $y \ge 0$.

Graphical method. Maximise $Z=x+y$ over $x+4y\le 8,\ 2x+3y\le 12,\ 3x+y\le 9,\ x,y\ge 0$.

Find the feasible corner points, testing each pairwise intersection against the third constraint:

• $(0,0)$, $(3,0)$ (from $3x+y=9$ on the $x$-axis), $(0,2)$ (from $x+4y=8$ on the $y$-axis).
• $x+4y=8 \cap 3x+y=9 \Rightarrow \left(\frac{28}{11},\frac{15}{11}\right)$; check $2x+3y=\frac{101}{11}\approx 9.18\le 12$ (ok). …

Idea

Each constraint is a half-plane; together they cut out a bounded polygon. By the corner-point theorem the maximum of the linear $Z=x+y$ sits at one of its vertices, so we list the vertices and compare.

Constraints

Find the vertices

Axes. On the $x$-axis the smallest intercept is from $3x+y=9$, giving $(3,0)$. On the $y$-axis the smallest is from $x+4y=8$, giving $(0,2)$. And the origin $(0,0)$.

$x+4y=8$ and $3x+y=9$. From the second, $y=9-3x$. Then $x+4(9-3x)=8 \Rightarrow -11x=-28 \Rightarrow x=\dfrac{28}{11},\ y=\dfrac{15}{11}$. Check the third constraint: $2x+3y=\dfrac{56+45}{11}=\dfrac{101}{11}\approx 9.18\le 12$ (ok) — feasible.

Reject the infeasible crossings.

• $x+4y=8$ with $2x+3y=12$ gives $\left(\dfrac{24}{5},\dfrac{4}{5}\right)$, but $3x+y=\dfrac{72+4}{5}=15.2>9$ — outside. …

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. …

Common Mistakes

Mistake 1: Using an infeasible intersection as a corner.

Why it's wrong: two of the pairwise crossings here, $\left(\tfrac{24}{5},\tfrac{4}{5}\right)$ and $\left(\tfrac{15}{7},\tfrac{18}{7}\right)$, violate the third constraint, so they are not vertices of the feasible region; evaluating $Z$ at them gives a false maximum. Correct approach: test every intersection against all three constraints and keep only those that satisfy them all. …