Q.Minimise $Z=3x+5y$ subject to $x+3y\ge 3,\ x+y\ge 2,\ x\ge 0,\ y\ge 0$.

The $\ge$ constraints give an unbounded region; find the corners, then apply the open-half-plane check.

Boundary lines (both $\ge$, so solid; the feasible side is away from the origin):

• $x+3y=3$: intercepts $(3,0),(0,1)$. At $(0,0)$: $0\ge3$ false — shade the side away from the origin.
• $x+y=2$: intercepts $(2,0),(0,2)$. At $(0,0)$: $0\ge2$ false — shade the side away from the origin.

With $x,y\ge0$, the feasible region is the unbounded region up-and-right of both lines, in the first quadrant.

Corner points:

• $(3,0)$: on $x+3y=3$; check $x+y=3\ge2$ ✓.
• Intersection of $x+3y=3$ and $x+y=2$: subtract: $(x+3y)-(x+y)=3-2\Rightarrow 2y=1\Rightarrow y=\tfrac12$, then $x=2-\tfrac12=\tfrac32$: $\left(\tfrac32,\tfrac12\right)$.
• $(0,2)$: on $x+y=2$; check $x+3y=6\ge3$ ✓.

Evaluate $Z=3x+5y$:

The smallest corner value is $7$. Because the region is unbounded, verify it is genuinely the minimum: draw the open half-plane $3x+5y<7$. This lies below-left of the line $3x+5y=7$, whereas the feasible region lies up-right of both boundary lines; the two have no point in common. Hence $Z$ cannot be made smaller than $7$, and $7$ is the true minimum. …

Skipping the open-half-plane check on an unbounded region and simply declaring the smallest corner value the minimum — on an unbounded region that step is what proves the minimum actually exists. Also, shading the o …