Q.Minimize $Z = 5x + 7y$ subject to the constraints $2x + y \ge 8,\ x + 2y \ge 10,\ x, y \ge 0$. Also determine, with reason, whether $Z$ has a maximum value on this feasible region.

Constraints: $2x+y\ge8,\ x+2y\ge10,\ x,y\ge0$.

Corner points. At $y=0$: first constraint needs $x\ge4$, second needs $x\ge10$; binding is $x\ge10$, giving vertex $(10,0)$ — check first: $20\ge8$, slack.

At $x=0$: first needs $y\ge8$, second needs $y\ge5$; binding is $y\ge8$, giving vertex $(0,8)$ — check second: $16\ge10$, slack.

Intersection: $2x+y=8$ and $x+2y=10$. From the first, $y=8-2x$; substituting, $x+2(8-2x)=10\Rightarrow x+16-4x=10\Rightarrow-3x=-6\Rightarrow x=2,\ y=4$, giving $(2,4)$.

Corner points: $(10,0),\ (2,4),\ (0,8)$ — and the region extends without bound away from the origin, along the positive $x$-direction beyond $(10,0)$ and along the positive $y$-direction beyond $(0,8)$.

Evaluating $Z=5x+7y$:

$(10,0)\to50,\qquad (2,4)\to10+28=38,\qquad (0,8)\to56.$ …

Find the corner points as usual, but for an unbounded region also check the open half-plane {Z < candidate} (for a minimum) or {Z > candidate} (for a maximum) against the feasible region — an empty intersection confirms that corner value is the genuine o …

Assuming the smallest corner value is automatically the minimum without applying the half-plane test, purely by analogy with the bounded case; wrongly claimin …