Maximize $Z=2x+3y$ subject to $x+y\le 5$, $x+2y\le 8$, $x,y\ge 0$.

Solution

1. The vertices are $(0,0)$, $(5,0)$, $(0,4)$, and the intersection of $x+y=5$ and $x+2y=8$.
2. Subtracting the equations: $$\begin{array}{rl}(x+2y)-(x+y)& =8-5⟹y\\ & =3\end{array}$$.
3. Substituting $y=3$ into $x+y=5$ gives $x=2$.
4. At $(2,3)$, $$\begin{array}{rl}Z& =2(2)+3(3)\\ & =4+9\\ & =13\end{array}$$.
5. At $(5,0)$, $Z=10$. At $(0,4)$, $Z=12$.
6. Hence the answer is (B).