Find the values of $ \ a,b \ $ so that the polyhedral set defined by the system of inequalities
$$ ax+(b+1)y \leq 120 \\ x+(a+b)y \leq 160 \\ (a-b)x+y \leq 30 \\ x \geq 0 \\ y \geq 0 $$
has extreme direction.
Answer:
Does extreme direction means that we have to solve the equality as follows:
$$ ax+(b+1)y = 120 \\ x+(a+b)y = 160 \\ (a-b)x+y = 30 \\ x \geq 0 \\ y \geq 0 $$
If so then how to solve the $ \ 3 \ $ above equations ?
Any hints will be helpful.
Solving for $x, y$
$$ ax+(b+1)y = 120\\ x+(a+b)y = 160 $$
giving
$$ x = \frac{40(3a-b-4)}{(a-1)(a+b+1)}\\ y = \frac{40(4a-3)}{(a-1)(a+b+1)} $$
Now substituting $x(a,b), y(a,b)$ into $(a-b)x+y=30$ we have
$$ f(a,b)= 4b^2+9a^2-19b(1-a)-9=0 $$
which describes the variety obeying extreme direction. Except for some values ...
Attached the plot showing $x(a,b) > 0$ and $y(a,b) > 0$ in light blue and $f(a,b) = 0$ in red