For the LPP, Min $z= 6x+10y $ subject to $x≥6, y ≥3, 2x+y ≥ 10, x ≥0, y≥0, $ redundant constraint is :

$2x+y ≥10$

The correct answer is Option (2) → $2x+y ≥10$

At $x≥6$, $y ≥3$ then,

$2x+y≥2(6)+3=15$

Since $15≥10$ always holds, the constraint

$2x+y≥10$

is automatically satisfied by any (x, y) that meets $x=6$ and $y≥3$

∴ The redundant constraint is

$2x+y ≥10$