Q.Show that $Z = 4x + 6y$ attains a minimum value, but no maximum value, on the feasible region determined by $x + y \ge 5,\ x + 2y \ge 8,\ x, y \ge 0$. Find the minimum value.

Constraints: $x+y\ge5,\ x+2y\ge8,\ x,y\ge0$.

At $y=0$: first needs $x\ge5$, second needs $x\ge8$; binding $x\ge8$, vertex $(8,0)$ — check first: $8\ge5$, slack.

At $x=0$: first needs $y\ge5$, second needs $y\ge4$; binding $y\ge5$, vertex $(0,5)$ — check second: $10\ge8$, slack.

Intersection: $x+y=5,\ x+2y=8$; subtracting, $y=3,\ x=2$, giving $(2,3)$.

Corners: $(8,0),\ (2,3),\ (0,5)$.

Minimum $=26$, at $(2,3)$. Because both coefficients of $Z$ are positive and the region is unbounded, moving out along either ray beyond $(8,0)$ or $(0,5)$ makes $Z$ grow without bound, so no maximum value exists.

Compute all corner points, evaluate Z, take the smallest as the minimum (valid for positive coefficients on an outward-receding region), and argue directly (not by trial) that no maximum can exist.

Trying to 'find the maximum' by checking only the given corner points and picking the largest (30), instead of recognizing that an unbounded region with positive-coefficient Z never has a maximum.