Kendrick Hampton

2022-06-25

Find the values of so that the polyhedral set defined by the system of inequalities
$ax+\left(b+1\right)y\le 120\phantom{\rule{0ex}{0ex}}x+\left(a+b\right)y\le 160\phantom{\rule{0ex}{0ex}}\left(a-b\right)x+y\le 30\phantom{\rule{0ex}{0ex}}x\ge 0\phantom{\rule{0ex}{0ex}}y\ge 0$
has extreme direction.

benedictazk

Expert

Solving for $x,y$
$ax+\left(b+1\right)y=120\phantom{\rule{0ex}{0ex}}x+\left(a+b\right)y=160$
giving
$x=\frac{40\left(3a-b-4\right)}{\left(a-1\right)\left(a+b+1\right)}\phantom{\rule{0ex}{0ex}}y=\frac{40\left(4a-3\right)}{\left(a-1\right)\left(a+b+1\right)}$
Now substituting $x\left(a,b\right),y\left(a,b\right)$ into $\left(a-b\right)x+y=30$ we have
$f\left(a,b\right)=4{b}^{2}+9{a}^{2}-19b\left(1-a\right)-9=0$
which describes the variety obeying extreme direction.

Do you have a similar question?