In Graphical solution the feasible solution is any solution to a … - Vidyadip

MCQ

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

A
Only objective function.
✓
Non $-$ negativity restriction.
C
Only constraint.
D
All the three

✓

Answer

Correct option: B.

Non $-$ negativity restriction.

The feasible region is the set of all the points that satisfy all the given constraints.
The variables of the linear programs must always take the non $-$ negative values $($i.e., $\text{x}\geq0$ and $\text{y}\geq0).$
These are used because x and y are usually the number of items produced and we cannot produce the negative number of items.
The least possible number of items could be zero.
Therefore, the feasible solution should satisfy the non $-$ negativity restriction.

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

If $m$ is chosen in the quadratic equation $\left( {{m^2} + 1} \right)\,{x^2} - 3x + {\left( {{m^2} + 1} \right)^2} = 0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is

The minimum value of $\frac{\text{x}}{\log_{\text{e}}\text{x}}$ is .

If $f(x)$ be a polynomial function satisfying $f(x).f (\frac{1}{x}) = f(x) + f (\frac{1}{x})$ and $f(4) = 65$ then value of $f(6)$ is

Consider the function $f : R \rightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\left(2-\sin \left(\frac{1}{x}\right)\right)|x|, x \neq 0 \\ 0 & , x=0\end{array} .\right.$ Then $f$ is

The minimum area of a triangle formed by any tangent to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{81}} = 1$ and the co-ordinate axes is

The reflection of the point $(\text{a}, \beta, \gamma) $ in the $xy-$plane is:

Let $f:[0, \infty) \rightarrow R$ be a continuous function such that $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ for all $x \in[0, \infty)$. Then, which of the following statement (s) is (are) $TRUE$?

$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$

$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$

$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$

$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$

The order of $[x\,y\,z]\,\,\left[ {\begin{array}{*{20}{c}}a&h&g\\h&b&f\\g&f&c\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right]$ is

The range of $a \in R$ for which the function $ f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ $x \neq 2 n \pi, n \in N ,$ has critical points, is

Let $x$ and $y$ be two positive real numbers such that $x+y=1$. Then, the minimum value of $\frac{1}{x}+\frac{1}{y}$ is