Q.Solve the LPP graphically: Maximise $Z=5x+4y$ subject to $2x+y\le 8,\ x+3y\le 9,\ x\ge 0,\ y\ge 0$.

Graph the two solid lines, shade the common region, then evaluate $Z$ at each corner.

Corners: $(0,0),(4,0),(3,2),(0,3)$; $Z=0,20,23,12$ respectively.

Step 1 — Boundary lines (all $\le$, so solid; test the origin).

• $2x+y=8$: intercepts $(4,0)$ and $(0,8)$. At $(0,0)$: $0\le8$ true — shade the origin side.
• $x+3y=9$: intercepts $(9,0)$ and $(0,3)$. At $(0,0)$: $0\le9$ true — shade the origin side.
• $x\ge0,\ y\ge0$: the first quadrant.

Step 2 — Feasible region. The overlap is a bounded quadrilateral in the first quadrant.

Step 3 — Corner points (solve boundaries in pairs):

• $x=0,\ y=0$: $(0,0)$.
• $y=0$ in $2x+y=8$: $x=4$, giving $(4,0)$ (check $x+3y=4\le9$ ✓).
• $x=0$ in $x+3y=9$: $y=3$, giving $(0,3)$ (check $2x+y=3\le8$ ✓).
• Intersection of $2x+y=8$ and $x+3y=9$: from the first $y=8-2x$; substitute: $x+3(8-2x)=9\Rightarrow x+24-6x=9\Rightarrow -5x=-15\Rightarrow x=3$, then $y=8-6=2$: the point $(3,2)$.

Step 4 — Evaluate $Z=5x+4y$ at each corner.

Step 5 — Choose the maximum. The largest value is $23$, at $(3,2)$. Since the region is bounded, this is the true maximum.

Check (independent verification): re-solve the intersection the other way — from $x+3y=9$, $x=9-3y$; substitute into $2x+y=8$: $2(9-3y)+y=8\Rightarrow 18-6y+y=8\Rightarrow -5y=-10\Rightarrow y=2$, $x=3$. Same point $(3,2)$, and $Z=5(3)+4(2)=23$ confirmed.

The iso-profit view: draw a line $5x+4y=k$ and slide it up-right; the last corner it touches inside the region is $(3,2)$, giving the maximum $k=23$ — the same answer as the corner table.

Reading the intersection off a rough sketch instead of solving the two equations, or swapping the intercepts of $2x+y=8$ (it meets the axes at $(4,0)$ and $(0,8)$, not $(8,0)$ and $(0,4)$).