Download App

Linear Programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear Programming1 Mark
Question
If the constraints in linear programming problem are changed.
✓

Answer

  1. The problem is to be re - evaluated
Solution:
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 re - evaluated.
In an 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 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→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?
  1. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
  2. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
  3. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
  4. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$ 
→
The values of the constants a, b and for which the function $\text{f(x)}=\begin{cases}(1+\text{ax})^{\frac{1}{\text{x}}},&\text{x}>0\\\text{b},&\text{x}=0\\\frac{(\text{x}+\text{c})^{\frac{1}{2}}-1}{(\text{x}+1)^{\frac{1}{2}}-1},&\text{x}>0\end{cases}$ may be continuous at x = 0, are:
  1. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=-\frac{2}{3},\text{ c}=1$
  2. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\frac{2}{3},\text{ c}=-1$
  3. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
  4. none of these
→
The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for $x = 2.$ in order to make $f(x)$ continuous at $x = 2,$ here $f(2)$ should be defined as:
→
If $R$ and $R^{\prime}$ are symmetric relations (not disjoint) on a set $A$, then the relation $R \cap R^{\prime}$ is
→
If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is equal to
→
Which one of the following function is not invertible?
  1. $\text{f} : \text{R} \rightarrow \text{R}, \text{f(x)} = 3\text{x} + 1$
  2. $\text{f} : \text{R} \rightarrow [0,\infty), \text{f(x)} = \text{x}^2$
  3. $\text{f} : \text{R}^+\rightarrow\text{R}^+, \text{f(x)} =\frac{1}{\text{x}^3}$
  4. None of these.
→
A bag contains six red four green and eight white balls If a ball is picked at random the probability that it is not white is:
  1. $\frac13$
  2. $\frac49$
  3. $\frac59$
  4. $\frac23$
→
If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{3}$
  3. $\frac{\pi}{2}$
  4. $-\frac{\pi}{3}$
→
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?
  1. 0, 1, 2
  2. 3, 5, 7
  3. 7, 7, 8
  4. 1, 5, 7
→
If the points $(p, 0), (0, q)$ and $(1, 1)$ are collinear, then $\frac{1}{\text{p}}+\frac{1}{\text{q}}$​ is equal to:
→