Q.Determine the feasible region of the system $2x+y\le 8,\ x+2y\le 10,\ x\ge 0,\ y\ge 0$ and list its corner points.

Shade all four half-planes; find the corners by solving boundary lines in pairs.

Boundary lines and shading (all $\le$, so all solid; origin as test point):

• $2x+y=8$: intercepts $(4,0),(0,8)$. At origin $0\le8$ true — shade the origin side.
• $x+2y=10$: intercepts $(10,0),(0,5)$. At origin $0\le10$ true — shade the origin side.
• $x\ge0,\ y\ge0$: the first quadrant.

The feasible region is where all four overlap — a bounded quadrilateral in the first quadrant.

Corner points (solve boundaries two at a time):

1. $x=0$ and $y=0$: $(0,0)$.
2. $y=0$ in $2x+y=8$: $2x=8\Rightarrow x=4$: $(4,0)$. (Check it also satisfies $x+2y=4\le10$ ✓.)
3. $x=0$ in $x+2y=10$: $2y=10\Rightarrow y=5$: $(0,5)$. (Check $2x+y=5\le8$ ✓.)
4. Intersection of the two slanted lines. From $2x+y=8$, $y=8-2x$. Substitute into $x+2y=10$:

then $y=8-2(2)=4$: the point $(2,4)$.

So the corners are $(0,0),(4,0),(2,4),(0,5)$. …

Reading the intersection of the two lines off a rough sketch instead of solving them algebraically; and swapping the intercepts of $2x+ …