A linear programming problem has constraints $x+y\le 1$, $3x+y\ge 3$, $x\ge 0$, $y\ge 0$. What is the nature of the feasible region?

Solution

1. The constraints are $x+y\le 1$, $3x+y\ge 3$, $x\ge 0$, and $y\ge 0$.
2. For the first constraint $x+y\le 1$ in the first quadrant, the maximum value of $3x + y$ is $3$ only at $(1,0)$, but at this point $x+y=1$.
3. For any point satisfying $x+y\le 1$ where $x,y\ge 0$, we have $3x+y=2x+(x+y)\le 2x+1$.
4. Since $x\le 1$ from the first constraint, $2x+1\le 3$, meaning $3x + y$ can only reach $3$ if $x=1$ and $y=0$, which contradicts $x+y\le 1$ unless $x+y=1$.
5. Testing the boundary $x+y=1$ shows that $3x+y = 2x+1$, which is less than $3$ for all $x<1$, so no points satisfy both inequalities simultaneously.
6. Hence the answer is (C).