Objective of LPP is: - Vidyadip

MCQ

Objective of LPP is:

A
A constraint
✓
A function to be optimized
C
A relation between the variables
D
None of the above

✓

Answer

Correct option: B.

A function to be optimized

The objective of Linear Programming Problems (LPP) is to minimize or maximize the function.

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 $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x}),$ then $y$ is a solution of:

Let $c, k \in R$. If $f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$ and $f(x+y)=f(x)+f(y)-x y$, for all $x, y \in R$, then the value of $|2( f (1)+ f (2)+ f (3)+\ldots \ldots+ f (20)) \mid$ is equal to

If $y = x + e^x$ , then at $x = 1$ , $\frac{{{d^2}x}}{{d{y^2}}}$ is equal to

The corner points of the feasible region determined by the system of linear constraints are $(0,10),(5,5),(15,15),(0,20) .$ Let $z=p x+q y,$ where $p, q\,>\,0 .$ Condition on $p$ and $q$ so that the maximum of $z$ occurs at both the pooints $(15,15)$ and $(0,20)$ is $\ldots \ldots$

If $A$ and $B$ are symmetric matrices of the same order, then:

The value of f(0), so that the function $\text{f(x)}=\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2},\text{ x}\neq0$ is continuous everywhere, is given by:

If $A = \left[ {\begin{array}{*{20}{c}}1&2&3\\1&4&9\\1&8&{27}\end{array}} \right]$, then the value of $|adj\,\,A|$is

If $x=t^2, y=t^3,$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2} =$

The ratio of the areas between the curves $\text{y}=\cos\text{x}$ and $\text{y}=\cos2\text{x}$ and $x-$ axis from $x = 0$ to $x = 0$ to $\text{x}=\frac{\pi}{3}$

Let the solution curve of the differential equation $x \frac{d y}{d x}-y=\sqrt{y^{2}+16 x^{2}}, y(1)=3$ be $y=y(x)$. Then $y (2)$ is equal to