Corner points of the bounded feasible region for an $LP$ problem are $(0,4), (6,0), (12,0), (12,16)$ and $(0,10)$. Let $z = 8x + 12y$ be the objective function. Match the following:
$(i)$ Minimum value of $z$ occurs at $\ldots$
$(ii)$ Maximum value of $z$ occurs at $\ldots$
$(iii)$ Maximum of $z$ is $\ldots$
$(iv)$ Minimum of $z$ is $\ldots$

  • A
    $(i) (6,0), (ii) (12,0), (iii) 288, (iv) 48$
  • B
    $(i) (0,4), (ii) (12,16), (iii) 288, (iv) 48$
  • C
    $(i) (0,4), (ii) (12,16), (iii) 288, (iv) 96$
  • D
    $(i) (6,0), (ii) (12,0), (iii) 288, (iv) 96$

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Similar Questions

The maximum value of $Z=x+3y$ subject to the constraints $2x+y \leq 20$,$x+2y \leq 20$,$x \geq 0$,$y \geq 0$ is:

The coordinates of the corner points of the bounded feasible region are $(0, 10)$, $(5, 5)$, $(15, 15)$, and $(0, 20)$. The minimum value of the objective function $z = 3x + 9y$ 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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View Solution

The corner points of the feasible region determined by a system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$. If the objective function is $z = px + qy$ where $p, q > 0$, then the condition on $p$ and $q$ such that the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$ is . . . . . . .

The corner points of the feasible region determined by $A (20, 10)$,$B (18, 12)$,and $C (12, 12)$. The maximum value of the objective function $Z = 2x + 3y$ is . . . . . . .

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