Download App

Linear Programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear Programming1 Mark
Question
In order for a linear programming problem to have a unique solution, the solution must exist.
  1. At the intersection of the nonnegativity constraints.
  2. At the intersection of a nonnegativity constraint and a resource constraint.
  3. At the intersection of the objective function and a constraint.
  4. At the intersection of two or more constraints.
  5. None of the above.
✓

Answer

  1. At the intersection of two or more constraints.
Solution:
In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints.
Then the problem becomes convex and has a single optimum (maximum or minimum).

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→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Let $f ( x )=[ x ]^2+\sqrt{x}$, where $[\bullet]$ and $[\bullet]$ respectively denotes the greatest integer and fractional part functions, then
→
If $y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)},$ then $\frac{d y}{d x}$ is equal to
→
Evaluate : $\int_0^2 e^{3-4 x} d x$
→
Find the cofactor of the element of third row and second column of the following determinant
$
\left|\begin{array}{lll}
1 & x & y+z \\
1 & y & z+x \\
1 & z & x+y
\end{array}\right| \text {. }
$
→
Choose the correct answer from the given four options.
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to:
  1. $\text{P}(\text{A})+\text{P}(\text{B})$
  2. $\text{P}(\text{A})-\text{P}(\text{B})$
  3. $\text{P}(\text{A})\cdot\text{P}(\text{B})$
  4. $\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
→
Let f : N → R : $\text{f}(\text{x})=\frac{(2\text{x}-1)}{2}$ and g : Q → R : g(x) = x + 2 be two functions. Then, (gof) $(\frac{3}{2})$ is:
  1. 3
  2. 1
  3. $\frac{7}{2}$
  4. None of these.
→
The value of a for which the function $\text{f(x)}=\begin{cases}5\text{x}-4,&\text{if }0<\text{x}\leq1\\4\text{x}^2+3\text{ax},&\text{if }<\text{x}<2\end{cases}$ is continuous at every point of its domain, is:
  1. $\frac{13}{3}$
  2. 1
  3. 0
  4. -1
→
The probability distribution of a discrete random variable $X$ is given below :
$X$0123
$P(X)$$\frac{4}{k}$$\frac{6}{k}$$\frac{4}{k}$$\frac{2}{k}$
The value of $k$ is
→
If matrix $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I,$ then $X=$
→
Find the area of the ellipse $\frac{x^2}{4^2}+\frac{y^2}{9^2}$.
→