Q.Solve graphically: Minimize $Z = 20x + 10y$ subject to $x + 2y \ge 40,\ 3x + y \ge 30,\ x, y \ge 0$. Does $Z$ have a maximum value on this feasible region? Justify your answer.

Constraints: $x+2y\ge40,\ 3x+y\ge30,\ x,y\ge0$.

At $y=0$: first needs $x\ge40$, second needs $x\ge10$; binding $x\ge40$, vertex $(40,0)$ — check second: $120\ge30$, slack.

At $x=0$: first needs $y\ge20$, second needs $y\ge30$; binding $y\ge30$, vertex $(0,30)$ — check first: $60\ge20$, slack.

Intersection: $x+2y=40,\ 3x+y=30$. From the second, $y=30-3x$; substituting, $x+2(30-3x)=40\Rightarrow x+60-6x=40\Rightarrow-5x=-20\Rightarrow x=4,\ y=30-12=18$, giving $(4,18)$.

Corners: $(40,0),\ (4,18),\ (0,30)$.

$Z=20x+10y:\quad 800,\ \ 80+180=260,\ \ 300.$ …

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