Q.Solve graphically: Maximize $Z = 5x + 4y$ subject to $x + 2y \le 8,\ 3x + 2y \le 12,\ x, y \ge 0$.

The feasible corners are $(3,0)$, $\left(\tfrac32,\tfrac12\right)$ and $(0,2)$; evaluating $Z=3x+5y$ gives $9,\,7,\,10$, so the minimum is $Z=7$ at $\left(\tfrac32,\tfrac12\right)$.

Concept. Solve graphically by identifying the corner points of the feasible region defined by the constraints, then apply the corner-point theorem. For an unbounded region we also confirm the objective cannot go lower than the best corner value. This is the standard NCERT Class 12 mathematics graphical LPP method.

Constraints. $x+3y\ge3$, $x+y\ge2$, $x,y\ge0$.

Corner points.

• Line $x+3y=3$ meets the $x$-axis at $(3,0)$; it satisfies $x+y\ge2$ ($3\ge2$) — feasible.
• Line $x+y=2$ meets the $y$-axis at $(0,2)$; it satisfies $x+3y\ge3$ ($6\ge3$) — feasible.
• Intersection of $x+3y=3$ and $x+y=2$: subtracting gives $2y=1\Rightarrow y=\tfrac12,\ x=\tfrac32$, i.e. $\left(\tfrac32,\tfrac12\right)$.

Evaluate $Z=3x+5y$.

$Z(3,0)=9,\qquad Z\!\left(\tfrac32,\tfrac12\right)=\tfrac92+\tfrac52=7,\qquad Z(0,2)=10.$

Unbounded check. The region is unbounded above/right, but since the coefficients of $Z$ are positive, $Z$ increases in those directions; the open half-plane $3x+5y<7$ contains no feasible point. Hence $7$ is the true minimum.