COMEDK 2025 Evening Shift | Linear Programming Question 2 | Mathematics

Given $Z=80 x+120 y$, subject to constraints are $x+3 y \leq 30 ; 3 x+4 y \leq 60 ; x \geq 0 ; y \geq 0$. P is one of the corner points of the feasible region for the given Linear Programming Problem. Then the coordinate of $P$ is

$(0,15)$

$(20,0)$

$(6,12)$

$(30,0)$

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

-0

The corner points of the feasible region determined by the system of linear constraints are $(0,3),(1,1)$ and $(3,0)$, If objective function is $Z=p x+q y, p, q>0$ then the condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3,0)$ and $(1,1)$ is

$p=3 q$

$3 p=q$

$p=\frac{q}{2}$

$p=2 q$

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

For a given Linear Programming problem, the objective function is

$$z=3 x+2 y$$

Subject to constraints are

$$\begin{aligned} & 4 x+3 y \leq 60 \\ & x \geq 3 \\ & y \leq 2 x \\ & y \geq 0 \end{aligned}$$

P is one of the corner points of the feasible region for the given Linear Programming problem. Then the coordinate of P is

$(3,6)$

$(0,20)$

$(0,0)$

$(12,6)$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

The maximum value of $$P=500 x+400 y$$ for the given constraints $$x+y \leq 200, \quad x \geq 20, \quad y \geq 4 x, \quad y \geq 0$$ is

96,000

84,000

98,000

82,000

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