For a linear programming problem, the objective function is $z = px + qy$, where $p, q > 0$. If at the corner points $(0, 10)$ and $(5, 5)$ the values of $z$ are $90$ and $60$ respectively, then the relation between $p$ and $q$ is . . . . . . .
- A$q = 3p$
- B$p = 3q$
- C$q = 2p$
- D$p = 2q$
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The feasible region for an $LPP$ is shown in the figure. Let $z=3x-4y$ be the objective function. The maximum value of $z$ is $....$
Medium
View SolutionThe corner points of the bounded feasible region are $(60,0), (120,0), (60,40), (40,20)$ and $(20,30)$. For the objective function $z=5x+10y$:
$(i)$ Maximum value of $z$.
$(ii)$ Minimum value of $z$.
$(iii)$ Maximum value of $z$ occurs at.
$(iv)$ Minimum value of $z$ occurs at.
$(i)$ Maximum value of $z$.
$(ii)$ Minimum value of $z$.
$(iii)$ Maximum value of $z$ occurs at.
$(iv)$ Minimum value of $z$ occurs at.
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View SolutionLet $x$ and $y$ be optimal solutions of a Linear Programming $(LP)$ problem. Then,which of the following is true?
Easy
View SolutionThe 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:
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:
$x + 2y \leq 10$
$3x + y \leq 15$
$x, y \geq 0$
Medium
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