CBSE Class 12 · 2024 · Linear ProgrammingLong Answer

Boundary lines: $x+2y=12$, $2x+y=12$ and $x+\frac{5}{4}y=5$ (i.e. $4x+5y=20$).
The feasible region is bounded with corner points:
$A(5,0)$, $B(6,0)$, $C(4,4)$, $D(0,6)$, $E(0,4)$,
where $C(4,4)$ is the intersection of $x+2y=12$ and $2x+y=12$.
Note that $Z=300x+600y=300(x+2y)$.
Evaluate $Z$:
$A(5,0)$: $Z=1500$
$B(6,0)$: $Z=1800$
$C(4,4)$: $Z=300(12)=3600$ ← maximum
$D(0,6)$: $Z=300(12)=3600$ ← maximum
$E(0,4)$: $Z=2400$
Since $Z$ attains the same value 3600 at the two adjacent corner points $C(4,4)$ and $D(0,6)$, the maximum occurs at every point on the segment $CD$ (which lies on $x+2y=12$) — the LPP has infinitely many optimal solutions.