Which of the following statements is correct? - Vidyadip

MCQ

Which of the following statements is correct?

A
Every $\text{LPP}$ admits an optimal solution
B
A $\text{LPP}$ admits unique optimal solution
✓
If a $\text{LPP}$ admits two optimal solution it has an infinite number of optimal solutions
D
The set of all feasible solutions of a $\text{LPP}$ is not a converse set

✓

Answer

Correct option: C.

If a $\text{LPP}$ admits two optimal solution it has an infinite number of optimal solutions

Optimal solution of $\text{LPP}$ has three types.
$1.$ Unique
$2.$ Infinite
$3.$ Does not exist.
Hence, it has infinite solution if it admits two optimal solution.

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 "MCQ" from Linear programming→Linear programming — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $|\bar{a}|=2,|\bar{b}|=3|\bar{c}|=4$ then $[\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}-\bar{a}]$ is equal to

$(\hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=$ _______

If $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ and $A \cdot \operatorname{adj} A=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$, then $k$ is equal to

If $\vec{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\vec{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of $(2 \vec{a}-\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}+2 \vec{b})]$ is

Distance of plane $x+y+z=3$ from origin is

The area bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is

$\int_0^{\frac{x}{4}} \sqrt{1+\sin 2 x} d x=$

Let X denote the number of hours that a bus travels in a city during a randomly selected day. The probability that X can take the value $x$ has the following form, where $k$ is some unknown constant.
$\mathrm{P}(\mathrm{X}=x)= \begin{cases}0.1, & \text { if } x=0 \\ \mathrm{kx}, & \text { if } x=1 \text { or } 2 \\ \mathrm{k}(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases}$
Then the value of $k$ is

$\bar{a}$ and $\bar{b}$ are two non-collinear vectors, then $x \overline{ a }+y \overline{b}$ (where $x$ and $y$ are scalars) represents a vector which is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+\lambda \hat{j}+\hat{k}, \vec{c}=\hat{i}-\hat{j}+4 \hat{k}$ and $\vec{a} \cdot(\vec{b} \times \vec{c})=10$, then $\lambda$ is equal to