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 minimum 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 following linear programming problem graphically:
Minimise $Z = 200x + 500y$.......$(1)$
subject to the constraints:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$
Minimise $Z = 200x + 500y$.......$(1)$
subject to the constraints:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$
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View SolutionThe 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 $(1,1)$.
For the linear programming constraints $x+2y \geq 10$,$3x+4y \leq 24$,$x \geq 0$,and $y \geq 0$,which of the following is not a corner point of the feasible region?
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View SolutionThe difference between the maximum value and the minimum value of the objective function $z = 3x + y$ subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, $y \geq 0$ is....
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