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 . . . . . . .
- A$p = 3q$
- B$3p = q$
- C$p = \frac{q}{2}$
- D$p = 2q$
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Minimise $Z = 3x + 2y$ subject to the constraints:
$x + y \geqslant 8$ ... $(1)$
$3x + 5y \leqslant 15$ ... $(2)$
$x \geqslant 0, y \geqslant 0$ ... $(3)$
$x + y \geqslant 8$ ... $(1)$
$3x + 5y \leqslant 15$ ... $(2)$
$x \geqslant 0, y \geqslant 0$ ... $(3)$
Medium
View SolutionThe corner points of the feasible region determined by the following system of linear inequalities: $2x + y \leq 10$,$x + 3y \leq 15$,$x, y \geq 0$ are $(0,0)$,$(5,0)$,$(3,4)$,and $(0,5)$. Let $Z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is:
Medium
View SolutionThe vertices of the feasible region determined by some linear constraints are $(0,2), (1,1), (3,3), (1,5)$. 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 $(3,3)$ and $(1,5)$ is . . . . . . .
The corner points of the feasible region determined by the system of linear constraints are $(2, 72)$,$(15, 20)$,and $(40, 15)$. Let $Z = 6x + 3y$ be the objective function. The minimum value of $Z$ occurs at:
The feasible solution for a $LPP$ is shown in the figure. Let $z=3x-4y$ be the objective function. The minimum value of $z$ occurs at:
Easy
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