For the constraints $x+y\le 5$, $x-y\ge 0$, $x\ge 0$, $y\ge 0$, what is the minimum value of $Z=x+2y$?

Solution

1. The feasible region is defined by the constraints $x+y\le 5$, $x-y\ge 0$, $x\ge 0$, and $y\ge 0$.
2. The vertices of this region are $(0,0)$, $(5,0)$, and $(2.5,2.5)$.
3. Evaluating the objective function $Z=x+2y$ at the origin $(0,0)$ gives $$\begin{array}{rl}Z& =0+2(0)\\ & =0\end{array}$$.
4. Since $x$ and $y$ are non-negative, $Z=x+2y$ cannot be less than $0$ within the feasible region.
5. Hence the answer is (A).