The optimal value of the objective function is attained at the po… - Vidyadip

MCQ

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

A
Given by intersection of inequation with y - axis only.
B
Given by intersection of inequation with x - axis only.
✓
Given by corner points of the feasible region.
D
None of these

✓

Answer

Correct option: C.

Given by corner points of the feasible region.

Given by 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

Explore more

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

Similar questions

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has

The points $A(4,\,5,\,1),B(0, - 1, - 1),C(3,\,9,\,4)$ and $D( - 4,\,4,\,4)$ are

The integrating factor of the differential equation $\left( {{x^2} - 1} \right)\frac{{dy}}{{dx}} + 2xy = x$ is

If the tangent to the curve $x = at^2, y = 2$ at is perpendicular to $x-$ axis, then its point of contact is :

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ - 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} - 2{x^2} - \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then

Let a function $f : R \rightarrow R$ is defined such that $3f(2x^2 -3x + 5) + 2f(3x^2 -2x + 4) = x^2 -7x + 9\ \ \ \forall x \in R$, then the value of $f(5)$ is-

If $\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$ constant, then the maximum value of $\operatorname{asin} \mathrm{x}+\mathrm{b} \cos \mathrm{x}$, is :

The angle between the lines whose direction cosines are proportional to $(1, 2, 1)$ and $(2, -3, 6)$ is

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {(1 + {y^2})} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive and integral, then $x$ and $y =$

If $\text{f(x)}=\begin{cases}\frac{1-\sin\text{x}}{(\pi-2\text{x}^2)}\times\frac{\log\sin\text{x}}{\log(1+\pi^2-4\pi\text{x}+4\text{x}^2)},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k =$