MCQ
In Graphical solution the redundant constraint is:
- AWhich forms the boundary of feasible region.
- BWhich do not optimizes the objective function.
- CWhich does not form boundary of feasible region.
- DWhich optimizes the objective function.
Solution:
A constraint in an LP model becomes redundant when the feasible region doesnt change by the removing the constraint.
For example, $2\text{x}+\text{y}\geq10$ and $6\text{x}+3\text{y}\geq30$ are constraints.
$6\text{x}+3\text{y}\geq30$
$\Rightarrow3\times(2\text{x}+\text{y})\geq3\times10$
$\Rightarrow2\text{x}+\text{y}\geq10$
which is same as the first constraint.
Therefore, $6\text{x}+3\text{y}\geq30$ can be removed.
By removing this constraint the feasible region doesnt change.
If the boundary of the feasible region is removed then feasible solution set changes.
Hence, redundant constraint cannot be the boundary of the feasible region.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The differential equation for which $\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}$ is a solution, is:
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}+\text{b})\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}-\text{b})\text{y}=0$
$x+y-z=2, x+2 y+\alpha z=1,2 x-y+z=\beta$. If the system has infinite solutions, then $\alpha+\beta$ is equal to $.....$