I'm trying to solve the following problem:
Minimize $x + y$ with the constraints
- $a x + y \leq 4$
- $1 \leq x \leq 3$
- $1 \leq y \leq 3$ where $a \in \mathbb{R}$.
Now, the first step is to determine the possible values for $a$. I simply drew the constraints in an xy-plane and found $a \leq 3$. Otherwise (i.e. for $a > 3$), the constraint-set would be empty.
Okay, now (i.e. for $a \leq 3$) the minimum has to be $2$ at the position $(x,y)=(1,1)$, since it's the smalled value possible w.r.t. the above constraints.
But this seems way too simple. What am I missing?
Maybe a proof rather than visualization helps.
If $a> 3$, suppose the feasible set is non-empty, then $x, y >0$, then we have $$a \le \frac{4-y}{x}\le \frac{4-1}{1}=3$$
which is a contradiction.
Hence for the feasible set to be non-empty, we have have $a \le 3$.
If $a \le 3$, at $(1,1)$, $ax+y=a+1 \le 4,$ hence $(1,1)$ is a feasible point.
Hence the optimal value is $2$ when $a \le 3$ and doesn't exist otherwise.