In solving the $LP$ problem: "Minimize $z = 6x + 10y$ subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$." The redundant constraints are $....$

  • A
    $x \geq 6, y \geq 2$
  • B
    $2x + y \geq 10, x \geq 0, y \geq 0$
  • C
    $x \geq 6$
  • D
    $x \geq 6, y \geq 0$

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Similar Questions

Maximize $Z=3x+4y$,subject to the constraints: $x+y \leq 1, x \geq 0, y \geq 0$.

The coordinates of the corner points of the bounded feasible region are $(0,10), (5,5), (15,15)$,and $(0,20)$. The maximum value of the objective function $Z = 10x + 20y$ is:

The corner points of the bounded feasible region are $(0,1), (0,7), (2,7), (6,3), (6,0), (1,0)$. For the objective function $Z = 3x - y$:
$(i)$ At which point is $Z$ minimum?
$(ii)$ At which point is $Z$ maximum?
$(iii)$ The maximum value of $Z$ is $\ldots$
$(iv)$ The minimum value of $Z$ is $\ldots$

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The corner points of the feasible region determined by the system of linear inequalities are $(0,3), (1,1)$ and $(3,0)$. Let $Z = px + qy$ where $p, q > 0$. Find the condition on $p$ and $q$ such that the minimum of $Z$ occurs at both $(3,0)$ and $(1,1)$.

$z = 30x - 30y + 1800$ is an objective function. The corner points of the feasible region are $(15, 0), (15, 15), (10, 20), (0, 20),$ and $(0, 15)$. $z$ has the minimum value at $\ldots$ point.

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