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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The constraints $-x_{1} + x_{2} \leq 1$,$-x_{1} + 3x_{2} \leq 9$,and $x_{1}, x_{2} \geq 0$ define:
The solution set of the constraints $x + 2y \geq 11$,$3x + 4y \leq 30$,$2x + 5y \leq 30$,$x \geq 0$,$y \geq 0$ includes the point:
Medium
View SolutionThe following graph represents a feasible region. The minimum value of $z = 5x + 4y$ is $\ldots \ldots$
Medium
View SolutionCorner 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:
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:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$
Medium
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