For a linear programming problem,the objective function is $Z = 10500x + 9000y$. If the corner points of the bounded feasible region are $(0,0)$,$(40,0)$,$(30,20)$,and $(0,50)$,then the maximum value of $Z$ is . . . . . . .
- A$5,96,000$
- B$4,95,000$
- C$6,20,000$
- D$4,50,000$
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Similar Questions
Solve the following Linear Programming Problem graphically:
Minimise $Z = x + 2y$
subject to the constraints:
$2x + y \geq 3$
$x + 2y \geq 6$
$x, y \geq 0$
Minimise $Z = x + 2y$
subject to the constraints:
$2x + y \geq 3$
$x + 2y \geq 6$
$x, y \geq 0$
Medium
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:
The feasible region for an $LPP$ is shown in the figure. Let $z = 3x - 4y$ be the objective function. The minimum value of $Z$ is:
Medium
View SolutionIf an $LPP$ admits an optimal solution at two consecutive vertices of a feasible region,then:
MediumKCET 2017
View SolutionMaximum value of $z = 3x + 4y$ subject to the constraints $x - y \leqslant -1$,$-x + y \leqslant 0$,and $x, y \geqslant 0$ is:
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