Minimize $Z=5x+7y$ subject to $2x+y\ge 8$, $x+2y\ge 10$, $x,y\ge 0$.

Solution

1. The lines are $2x + y = 8$ and $x+2y=10$.
2. Solving the system: $y=8-2x$, so $$\begin{array}{rl}x+2(8-2x)& =10⟹x+16-4x\\ & =10⟹3x\\ & =6⟹x\\ & =2,y\\ & =4\end{array}$$.
3. The vertices of the unbounded region are $(8,0)$, $(2,4)$, and $(0,5)$.
4. At $(8,0)$, $Z=40$. At $(2,4)$, $$\begin{array}{rl}Z& =5(2)+7(4)\\ & =10+28\\ & =38\end{array}$$. At $(0,5)$, $Z=35$.
5. Re-evaluating: $2x+y=8$ and $x+2y=10$. $$\begin{array}{rl}x& =2,y\\ & =4\end{array}$$ gives $Z=38$. $$\begin{array}{rl}x& =8,y\\ & =0\end{array}$$ gives $Z=40$. $$\begin{array}{rl}x& =0,y\\ & =5\end{array}$$ gives $Z=35$.
6. Hence the answer is (B).