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

The area of region bounded by curve  $\text{y}=\cos2\text{x},$ line $\text{x}=0$ $\text{x}=\frac{\pi}{3}$ is:
  1. $\frac{2-\sqrt{3}}{4}$
  2. $\frac{\sqrt{3}}{4}$
  3. $\frac{4-\sqrt{3}}{4}$
  4. $\frac{\sqrt{3}-4}{4}$
If $\vec a\, = \,\vec i - 2\hat j + 3\hat k,\,\,\,\vec b = 2\vec i + 3\hat j - \hat k$ and $\vec c = \lambda \vec i + \hat j + (2\lambda  - 1\hat k)$ are coplanar vectors, then $\lambda $ is equal to
Let $f(x) = \left| {\begin{array}{*{20}{c}}{\,\sec x}&{\cos x}&{{{\sec }^2}x + \cot x\,{\rm{cosec}}\,x\,}\\{{{\cos }^2}x}&{{{\cos }^2}x}&{{\rm{cose}}{{\rm{c}}^2}x}\\1&{{{\cos }^2}x}&{{{\cos }^2}x}\end{array}} \right|\,,$ then $\int_0^{\pi /2} {\,f(x)\,dx = } $
Let × be a binary operation on set of integers I, defined by a × b = a + b - 3, then find the value of 3 × 4.
  1. 2
  2. 4
  3. 7
  4. 6
If $y = \sqrt {{{1 + {e^x}} \over {1 - {e^x}}}} $, then ${{dy} \over {dx}} = $
If $a = 2i + 4j - 5k$ and $b = i + 2j + 3k$, then $|a \times b|$ is
Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.
  1. $\frac{1}{2}$
  2. $\frac{1}{3}$
  3. $\frac{2}{3}$
  4. $\frac{4}{7}$
Given that $\left[\begin{array}{ll}1 & x\end{array}\right]\left[\begin{array}{cc}4 & 0 \\ -2 & 0\end{array}\right]=0$, the value of $x$ is :
Let $f(x)=a x^{2}+b x+c$ be such that $f(1)=3, f(-2)$ $=\lambda$ and $f (3)=4$. If $f (0)+ f (1)+ f (-2)+ f (3)=14$, then $\lambda$ is equal to$...$
Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors, such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 k$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to.