The corner points of the feasible region determined by $A (20, 10)$,$B (18, 12)$,and $C (12, 12)$. The maximum value of the objective function $Z = 2x + 3y$ is . . . . . . .
- A$72$
- B$82$
- C$80$
- D$70$
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Similar Questions
The corner points of the bounded feasible region are $(0,1), (0,7), (2,7), (6,3), (6,0), (1,0)$. For the objective function $Z = 3x - y$:
$(i)$ At which point is $Z$ minimum?
$(ii)$ At which point is $Z$ maximum?
$(iii)$ The maximum value of $Z$ is $\ldots$
$(iv)$ The minimum value of $Z$ is $\ldots$
$(i)$ At which point is $Z$ minimum?
$(ii)$ At which point is $Z$ maximum?
$(iii)$ The maximum value of $Z$ is $\ldots$
$(iv)$ The minimum value of $Z$ is $\ldots$
Difficult
View SolutionIf for a linear programming problem the feasible region is bounded,then the objective function has . . . . . . .
The feasible region for a $LPP$ is shown in the figure. Find the minimum value of $Z=11x+7y$.
Medium
View SolutionThe coordinates of the corner points of the bounded feasible region are $(0, 10)$, $(5, 5)$, $(15, 15)$, and $(0, 20)$. The minimum value of the objective function $z = 3x + 9y$ is . . . . . . .
MediumGUJCET 2026
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
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