Max Z= $\alpha \cdot x_1$ + $(1-\alpha)\cdot x_2$
(1) $2x_1$ + $x_2$ <= $4$
(2) $-2x_1$ + $x_2$ >= $0$
(3) $x_2$ >= 1
$x_1$ $>=$ 0
Given the following optimal solution polygon:
for which $\alpha$ values, (1,2) is an optimal solution?
a. $2/3 <\alpha<2$
b. $\alpha>2$
c. $\alpha<2/3$
I thought that the right answer is c because as long as I take smaller $\alpha$ I get higher value, but it seems to be the wrong answer.
The $4$ vertices are $(0,1)$, $(0.5, 1)$, $(1,2)$, $(0,4)$.
The corresponding objective values are $1-\alpha$, $0.5\alpha + (1-\alpha)=1-0.5\alpha$, $\alpha +2(1-\alpha)=2-\alpha$, $4(1-\alpha)=4-4\alpha$.
Let's study the question when does $2-\alpha$ becomes larger than the rest,
$2-\alpha \ge 1-\alpha$ is trivially true.
$2-\alpha \ge 1-0.5\alpha$ is equivalent to $1 \ge 0.5\alpha$, hence we have $2 \ge \alpha$.
For $2-\alpha \ge 4 - 4\alpha$, I would like to leave it as an exercise for you.