Question
The optimal value of the objective function is attained at the points

Answer

(c) : When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at corner points of the 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

If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$
  1. -2
  2. -1
  3. 1
  4. 2
The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 - b^2| < 16\}$ is given by$:$
A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is
The value of $\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$ is:
  1. $\frac{\pi}{2}$
  2. $\frac{5\pi}{3}$
  3. $\frac{10\pi}{3}$
  4. $0$
The value of the determinant $\left|\begin{array}{ccc}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right|$ is
Let $f: R \rightarrow R$ be defined by $f(x)=x+|x|$. Then $f(x)$ is
The differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$ has the general solution:
$\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]$ is equal to
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, then $\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$ equals:
  1. $0$
  2. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  3. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  4. $-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
A plane passing through (−1, 2, 3) and whose normal makes equal angle with the coordinate axes is:
  1. x + y + z + 4 = 0
  2. x − y + z + 4 = 0
  3. x + y + z − 4 = 0
  4. x + y + z = 0