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 $(1,1)$.
- A$p = 2q$
- B$p = \frac{q}{2}$
- C$p = 3q$
- D$p = q$
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The 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 . . . . . . .
Solve the following Linear Programming Problem graphically:
Maximise $Z = 3x + 2y$
subject to the constraints:
$x + 2y \leq 10$
$3x + y \leq 15$
$x, y \geq 0$
Maximise $Z = 3x + 2y$
subject to the constraints:
$x + 2y \leq 10$
$3x + y \leq 15$
$x, y \geq 0$
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View SolutionThe feasible region for a $LPP$ is shown in the figure. Find the maximum value of $Z=11x+7y$.
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