The corner points of the feasible region determined by the system of linear constraints are $(0,10), (10,15), (15,25), (0,30)$. Let $z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $z$ occurs at both the points $(15,25)$ and $(0,30)$ is . . . . . . .
- A$p:q = 2:1$
- B$p:q = 1:1$
- C$p:q = 2:3$
- D$p:q = 1:3$
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Which of the following terms is not used in a linear programming problem?
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 . . . . . . .
The 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.
Difficult
View SolutionSolve the Linear Programming Problem graphically:
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$
Medium
View SolutionDetermine graphically the minimum value of the objective function
$Z = -50x + 20y$ .....$(1)$
subject to the constraints:
${2x - y \geqslant -5}$ .....$(2)$
${3x + y \geqslant 3}$ .....$(3)$
${2x - 3y \leqslant 12}$ .....$(4)$
${x \geqslant 0, y \geqslant 0}$ .....$(5)$
$Z = -50x + 20y$ .....$(1)$
subject to the constraints:
${2x - y \geqslant -5}$ .....$(2)$
${3x + y \geqslant 3}$ .....$(3)$
${2x - 3y \leqslant 12}$ .....$(4)$
${x \geqslant 0, y \geqslant 0}$ .....$(5)$
Medium
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