Download App

Linear Programming — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsLinear Programming1 Mark
MCQ
If the constraints in linear programming problem are changed.
  • ✓
    The problem is to be $\text{re }- $ evaluated
  • B
    Solution is not defined
  • C
    The objective function has to be modified
  • D
    The change in constraints is ignored
✓

Answer

Correct option: A.
The problem is to be $\text{re }- $ evaluated
The above question asks for the impact of change in constraints on the Linear programming problem.
In this scenario, when there is a change in constraint, the solution will change definitely.
Whether the solution exists or not, we can only find once the problem is $\text{re} -$ evaluated.
In an $\text{LPP},$ the objective function is related to the main objective of any problem, either we have to maximize or minimize the function based on the situation whereas the constraints is related to physical restrictions in achieving the defined objective function.
In real life problems, there might be situations when the constraints change, but objective function does not changes to accommodate the change in constraints.
Thus, if constraints in linear programming problem is changed, the problem has to be $\text{re} -$ evaluated for the same objective function and after solving we can find whether the solution exists or not.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Linear Programming→Linear Programming — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is
→
$\cos^{-1}\Big(\frac{1}{2}\Big)$
→
The solution curve of the differential equation, $\left(1+ e ^{- x }\right)\left(1+ y ^{2}\right) \frac{ dy }{ dx }= y ^{2},$ which passes through the point $(0,1),$ is
→
Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is
→
$\begin{bmatrix}2\text{x}&5\\8&\text{x}\end{bmatrix}=\begin{bmatrix}6\text{x}&-2\\7&3\end{bmatrix},$ then the value of $\text{x}$ is:
→
For the given half-cylinder of volume $V$, the total surface area $S$ is minimum, when
→
If $y\sec x + \tan x + {x^2}y = 0$, then ${{dy} \over {dx}} =$
→
The value of $\int\limits_{ - 7}^7 {\frac{{{5^x}}}{{{5^{[x]}}}}dx} $ is equal to (where $[.]$ denotes greatest integer function)
→
Set $A$ has $3$ elements, and set $B$ has $4$ elements. Then the number of injective mappings that can be defined from $A$ to $B$ is:
→
The sum of the squares of sine of the angles made by the line $AB$ with $\text{OX, OY, OZ}$ where $O$ is the origin is:
→