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

Similar questions

In which interval $f(x) = 2x^2 - \ln |x| ,$  $(x \ne 0)$ is monotonically decreasing -
$\int_a^b {\frac{{\log x}}{x}\,dx = } $
Let $u,\,v,\,w$ be such that $|u|\, = 1,\,|v|\, = 2,\,|w|\, = 3.$ If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v,\,\,w$ are perpendicular to each other then $|u - v + w|$ equals
If $g(1) = g(2)$ then the value of $\int\limits_1^2 {{{[f\{ g(x)\} ]}^{ - 1}}f'\{ g(x)\} g'(x)dx} $ is-
If $f(x) = \,|x|,$ then $f'(0) = $
For a polynomial $g ( x )$ with real coefficient, let $m _{ g }$ denote the number of distinct real roots of $g ( x )$. Suppose $S$ is the set of polynomials with real coefficient defined by

$S=\left\{\left(x^2-1\right)^2\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right): a_0, a_1, a_2, a_3 \in R\right\} \text {. }$

For a polynomial $f$, let $f^{\prime}$ and $f^{\prime \prime}$ denote its first and second order derivatives, respectively. Then the minimum possible value of $\left(m_f+m_{f^{\prime}}\right)$, where $f \in S$, is. . . . . . . .

If $\text{y}^2=\text{ax}^2+\text{bx}+\text{c},$ then $\text{y}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is:
  1. a constant
  2. a function of x only
  3. a function of y only
  4. a function of x and only
Let $f :[0,1] \rightarrow R$ be a twice differentiable function in $(0,1)$ such that $f (0)=3$ and $f (1)=5$. If the line $y=2 x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$, then the least number of points $x \in(0,1)$, at which $f ^{\prime \prime}( x )=0$, is$......$
The area of a parallelogram whose two adjacent sides are represented by the vector $3i - k$ and $i + 2j$ is
For $\alpha, \beta, \gamma \neq 0$. If $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$, then $\gamma$ equal to