In linear programming, objective function and objective constrain… - Vidyadip

MCQ

In linear programming, objective function and objective constraints are:

A
Solved
✓
Linear
C
Quadratic
D
Adjacent

✓

Answer

Correct option: B.

Linear

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\text{A}=\displaystyle \begin{vmatrix} 2 &\text{amp;}-3 \\ 3 &\text{amp; 2} \end{vmatrix}$ and $\text{B}=\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}$ then $2\text{ A-B}=$

$\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)=$

Let the unit vectors $ a$ and $b$ be perpendicular and the unit vector $ c $ be inclined at an angle $\theta$ to both $a$ and $ b$ . If $c = \alpha \,a + \beta \,b + \gamma \,(a \times b),$ then

$f : R \rightarrow R$ is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is:

If $u = {\tan ^{ - 1}}(x + y),$ then $x{{\partial u} \over {\partial x}} + y{{\partial u} \over {\partial y}} = $

$A$ and $B$ are two square matrix such that $A^2B = BA$ and If $(AB)^{10} = A^K B^{10}$ then $k$ is

For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, b ∈ R − {1}, the inverse of a is:

Let the value of $p = \,\left( {x + 4y} \right)\vec a\, + \,\left( {2x + y + 1} \right)\vec b\,and\,q = \,\left( {y - 2x + 2} \right)\vec a\, + \,\left( {2x - 3y - 1} \right)\vec b\,,$ where $\vec a\,and\,\vec b$ non-collinear vectors. If $3p=2q,$ then the value of $x$ and $y$ will be

If the random variable X has the following distribution:

X:	0	1	2	3	4	5	6	7	8
P(X):	a	3a	5a	7a	9a	11a	13a	15a	17a

then the value of a is:

If the line joining $(2,3,-1)$ and $(3,5,-3)$ is perpendicular to the line joining $(1,2,3)$ and $(3,5, \lambda)$, then $\lambda=$