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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.

MAS
  • 10,638

1 Answers1

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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 ...

enter image description here

Attached the plot showing $x(a,b) > 0$ and $y(a,b) > 0$ in light blue and $f(a,b) = 0$ in red

Cesareo
  • 33,252
  • what are the values of $ \ a , b \ $ ? – MAS May 02 '18 at 15:00
  • Do we have to maximize or minimize $ \ f(a,b) \ $ in order to have extreme direction ? And there we will get values of $ \ a, b \ $ – MAS May 02 '18 at 15:02
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    All the pairs $(a,b)$ obeying $f(a,b) = 0$ and internal to the light blue region give extreme directions. – Cesareo May 02 '18 at 15:22