Question

The value of objective function is maximum under linear constraints

  1. at the centre of feasible region
  2. at (0, 0)
  3. at any vertex of feasible region
  4. the vertex which is maximum distance from (0, 0)

Answer

  1. at any vertex of feasible region

Solution:

In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.

Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.

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

Given that A is a square matrix of order 3 and |A| = -4, then |adj A| is equal to:
  1. -4
  2. 4
  3. -16
  4. 16
If $\text{f}(\text{x})=\tan^{-1}(\text{g}(\text{x})),$ where g(x) is monotonically increasing for $0<\text{x}<\frac{\pi}{2}.$ Then, f(x) is:
  1. Increasing on $\Big(0,\frac{\pi}{2}\Big)$
  2. Decreasing on $\Big(0,\frac{\pi}{2}\Big)$
  3. Increasing on $\Big(0,\frac{\pi}{4}\Big)$ and decreasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
  4. None of these.
Let $A=\{0,3,4,6,7,8,9,10\} \quad$ and $R$ be the relation defined on A such that $R =\{( x , y ) \in A \times A : x - y \quad$ is odd positive integer or $x-y=2\}$. The minimum number of elements that must be added to the relation $R$, so that it is a symmetric relation, is equal to $...........$.
If $\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$ is expressed as the sum of a symmetric and skew-symmetric matrix, then the symmetric matrix is:
  1. $\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$
  2. $\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$
  3. $\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$
  4. $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
If $[.]$ represents the greatest integer function, then the value of $\int_{0}^{\sqrt{\pi / 2}}\left(\left[ x ^{2}\right]+[-\cos x ]\right) d x$ is.............
The value of the function $(x - 1){(x - 2)^2}$ at its maxima is
Choose the correct option from given four options:
$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:
  1. $\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
  2. $\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
  3. $\frac{1}{10\text{x}}(1+4)^{-5}+\text{C}$
  4. $\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
The solution of the differential equation $\frac{{dy}}{{dx}} = 1 + x + y + xy$ is
Given that $f (x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f (a) < 0$ and $f (b) > 0$, which of the following are always true ?

$(i)$ $f (x)$ is bounded on $a \le x \le b.$

$(ii)$ The equation $f (x) = 0$ has at least one solution in $a < x < b.$

$(iii)$ The maximum and minimum values of $f (x)$ on $a \le x \le b$ occur at points where $f ' (c) = 0$.

$(iv)$ There is at least one point $c$ with $a < c < b$ where $f ' (c) > 0$.

$(v)$ There is at least one point $d$ with $a < d < b$ where $f ' (c) < 0.$

If the projection of $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ on $\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$ is zero, then the value of $\lambda$ is:
  1. $0$
  2. $1$
  3. $\frac{-2}{3}$
  4. $\frac{-3}{2}$