Question
In transportation models designed in linear programming, points of demand is classified as:

Answer

  1. Destinations
Solution:
In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes.
i.e., how effectively the available resources are transported to different destinations with minimum cost.
Therefore, the points of demand is classified as destinations.

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

$\mid\text{a}\times\text{b}\mid^2+\mid\text{a.b}\mid^2=144$ and $\mid\text{a}\mid=4$ then $\mid\text{b}\mid$ is equal to:
  1. 12
  2. 3
  3. 8
  4. 4
If $4a + 2b + c = 0,$ then the equation $3ax^2+ 2bx + c = 0$ has atleast one real root lying in the interval:
The equation of tangent to the curve $y (1 + x^2) = 2 - x,$ where it crosses $x-$axis is:
If order of matrix $A$ is $2 \times 3$, of matrix $B$ is $3 \times 2$, and of matrix $C$ is $3 \times 3$, then which one of the following is not defined?
Choose the correct answer from the given four options.Let f : N → R be the function defined by $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g : Q → R be another function defined by g(x) = x + 2. Then $(\text{gof})\frac{3}{2}$ is:
  1. $1$
  2. $1$
  3. $\frac{7}{2}$
  4. $\text{None of these}.$
If $\vec{a}, \vec{b}$ and $(\vec{a}+\vec{b})$ are all unit vectors and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the value of $\theta$ is:
The value of $|A|$, if $A=\left[\begin{array}{ccc}0 & 2 x-1 & \sqrt{x} \\ 1-2 x & 0 & 2 \sqrt{x} \\ -\sqrt{x} & -2 \sqrt{x} & 0\end{array}\right]$, where $x \in R ^{+}$, is
$\cot ^{-1} \frac{\sqrt{1-x^2}}{x}$ equal to :
If $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{J}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix},$ then B equals:
  1. $\text{I}\cos\theta+\text{J}\sin\theta$
  2. $\text{I}\sin\theta+\text{J}\cos\theta$
  3. $\text{I}\cos\theta-\text{J}\sin\theta$
  4. $-\text{I}\cos\theta+\text{J}\sin\theta$
The equation of the plane passing through (2, −3, 1) and is normal to the line joining the points (3, 4, −1) and (2, −1, 5) is given by:
  1. x + 5y − 6z + 19 = 0
  2. x − 5y + 6z − 19 = 0
  3. x + 5y + 6z + 19 = 0
  4. x − 5y − 6z − 19 = 0