Download App

Linear Programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear Programming1 Mark
Question
In linear programming, objective function and objective constraints are:
  1. Solved
  2. Linear
  3. Quadratic
  4. Adjacent
✓

Answer

  1. Linear
Solution:
In linear programming, objective function and objective constraints are linear.
Any linear programming problem must have the following properties:-1.
The relationship between variables and constraints must be linear 2.
The constraints must be non - negative.3.. objective function must be linear.

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

A coin is tossed 10 times. The probability of getting exactly six heads is:
  1. $\frac{512}{513}$
  2. $\frac{105}{512}$
  3. $\frac{100}{153}$
  4. $\text{ }^{10}\text{C}_6$
→
The orthogonal projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is:
  1. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
  2. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
  3. $\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
  4. $\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
→
Every invertible function is:
  1. Monotonic function.
  2. Constant function.
  3. Identity function.
  4. Not necessarily monotonic function.
→
A bag containe 5 black, 4white balls and 3 red balls. if a ball is selected randomwise, the probability that it is black or red ball is,
  1. $\frac{1}{3}$
  2. $\frac{1}{4}$
  3. $\frac{5}{12}$
  4. $\frac{2}{3}$
→
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
→
$S$ is a relation over the set $R$ of all real numbers and it is given by $(\text{a, b})\in\text{S}\Leftrightarrow\text{ab}\geq0.$ Then, $S$ is:
→
A speaks truth in 75% cases and B seaks truth in 80% cases. Probability that they contradict each other in a statement, is
  1. $\frac{7}{20}$
  2. $\frac{13}{20}$
  3. $\frac{3}{5}$
  4. $\frac{2}{5}$
→
What is the order of differential equation y’’ + 5y’ + 6 = 0?
  1. 0
  2. 1
  3. 2
  4. 3
→
The area bounded by display style $y = xe^{∣X∣} $ and $|x| = 1$ is:
→
The differential equation whose solution is $y=A e^{3 x}+B e^{-3 x}$ is given by
→