Q.Solve graphically: Minimize $Z = 3x + 4y$ subject to $x + 2y \ge 8,\ 3x + 2y \ge 12,\ x, y \ge 0$. State, with reason, whether $Z$ also has a maximum value on this region.

Constraints: $x+2y\ge8,\ 3x+2y\ge12,\ x,y\ge0$.

At $y=0$: first needs $x\ge8$, second needs $x\ge4$; binding $x\ge8$, vertex $(8,0)$ — check second: $24\ge12$, slack.

At $x=0$: first needs $y\ge4$, second needs $y\ge6$; binding $y\ge6$, vertex $(0,6)$ — check first: $12\ge8$, slack.

Intersection: $x+2y=8,\ 3x+2y=12$; subtracting, $2x=4\Rightarrow x=2,\ y=3$, giving $(2,3)$.

Corners: $(8,0),\ (2,3),\ (0,6)$.

Minimum $=18$, at $(2,3)$ (both coefficients of $Z$ positive, region recedes outward, so this is a genuine minimum). Since the region is unbounded and $Z$'s coefficients are positive, $Z\to\infty$ as $x\to\infty$ along $y=0$ or as $y\to\infty$ along $x=0$, so no maximum value exists.

Find corners as usual; for the maximum question on an unbounded region with positive objective coefficients, reason directly that Z grows without bound along the unbounded rays rather than searching for a largest corner value.

Reporting the larger of the two tied corner values (24) as 'the maximum', when in fact neither corner is a maximum — the region is unbounded and Z keeps growing beyond both.