The value of objective function is maximum under linear constrain… - Vidyadip

MCQ

The value of objective function is maximum under linear constraints

A
at (0, 0)
✓
at any vertex of feasible region
C
the vertex which is maximum distance from (0,0)
D
at the centre of feasible region

✓

Answer

Correct option: B.

at any vertex of feasible region

(b) at any vertex of feasible region
Explanation: 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

Explore more

More "M.C.Q (1 Marks)" from Model Paper 8→Model Paper 8 — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The angle between the curves $y^2=x$ and $x^2=y$ at $(1,1)$ is:

The transpose of a row matrix is :

$l,m,n$ are the ${p^{th}},{q^{th}}$and ${r^{th}}$term of a G.P., all positive, then $\left| {\,\begin{array}{*{20}{c}}{\log l}&{p\,\,\,\,\,\begin{array}{*{20}{c}}1\end{array}}\\{\log m}&{q\,\,\,\,\,\begin{array}{*{20}{c}}1\end{array}}\\{\log n}&{r\,\,\,\,\,\begin{array}{*{20}{c}}1\end{array}}\end{array}\,} \right|$ equals

Two cards are drawn at random and without replacement from a pack of 52 playing cards, then the probability that both the cards are black is :

What is the solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=0?$ Where c is a constant.

The shortest distance between the lines $\frac{{x - 3}}{3} = \frac{{y - 8}}{{ - 1}} = \frac{{z - 3}}{1}\& \frac{{x + 3}}{{ - 3}} = \frac{{y + 7}}{2} = \frac{{z - 6}}{4}$ is

The differential equation of the family of curves ${y^2} = 4a(x + a)$, where $a$ is an arbitrary constant, is

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ then the value of $|\vec{a}|^2$ will be $-$

Let $A =\{1,2,3,4, \ldots .10\}$ and $B =\{0,1,2,3,4\}$ The number of elements in the relation $R =\{( a , b )$ $\left.\in A \times A : 2( a - b )^2+3( a - b ) \in B \right\}$ is $.........$.

In Graphical solution the feasible solution is any solution to a $\text{LPP}$ which satisfies.