The corner points of the feasible region determined by the 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 $Z = qx + py$ where $p, q > 0$. The condition on $p$ and $q$ such that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is:
- A$p = 3q$
- B$2q = 3p$
- C$q = 3p$
- D$2p = 3q$
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Which of the following statements is correct?
Easy
View SolutionSolve the Linear Programming Problem graphically:
Maximise $Z = 5x + 3y$
subject to the constraints:
$3x + 5y \leq 15$
$5x + 2y \leq 10$
$x \geq 0, y \geq 0$
Maximise $Z = 5x + 3y$
subject to the constraints:
$3x + 5y \leq 15$
$5x + 2y \leq 10$
$x \geq 0, y \geq 0$
Medium
View SolutionFor a linear programming problem,the objective function is $Z = 3x + 2y$. If the corner points of the bounded feasible region are $(12, 0)$,$(4, 2)$,$(1, 5)$,and $(1, 10)$,then the maximum value of $Z$ is . . . . . . .
For the objective function $Z = 4x + y$ subject to the constraints $x + y \leq 50$,$3x + y \leq 90$,$x \geq 0$,$y \geq 0$,whose corner points of the feasible region are $(0,0)$,$(30,0)$,$(20,30)$,and $(0,50)$,the maximum value of $Z$ 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:
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