Corner points of the feasible region determined by the system of linear constraints are $(0,3), (1,1)$ and $(3,0)$. Let $Z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ such that the maximum of $Z$ occurs at both $(3,0)$ and $(1,1)$ is $.....$
- A$p = 2q$
- B$p = \frac{q}{2}$
- C$p = 3q$
- D$p = q$
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Solve the Linear Programming Problem graphically:
Maximise $Z = 3x + 4y$
subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0.$
Maximise $Z = 3x + 4y$
subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0.$
Medium
View SolutionThe corner points of the feasible region determined by the system of linear constraints are $(0,10), (10,15), (15,25), (0,30)$. Let $z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $z$ occurs at both the points $(15,25)$ and $(0,30)$ is . . . . . . .
For the $LP$ problem,minimize $z = 2x + 3y$,the coordinates of the corner points of the bounded feasible region are $A(3, 3), B(20, 3), C(20, 10), D(18, 12),$ and $E(12, 12)$. The minimum value of $z$ is:
Medium
View SolutionThe corner points of the feasible region determined by the system of linear constraints are $(0, 10)$, $(5, 5)$, $(15, 15)$, and $(0, 20)$. Let $z = px + qy$ where $p, q > 0$. The condition on $p$ and $q$ such that the maximum value of $z$ occurs at both points $(15, 15)$ and $(0, 20)$ is
DifficultKCET 2026
View SolutionThe minimum value of $Z = 2x + 3y$ for the system of linear constraints: $2x + 4y \leq 12$,$x + y \leq 3$,$x \geq 0$,and $y \geq 0$ is . . . . . . .
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