- given by intersection of inequations with the axes only
- given by intersection of inequations with x-axis only
- given by corner points of the feasible region
- none of these
Solution:
It is known that the optimal value of the objective function is attained at any of the corner point.
Thus, the potimal value of the objective function is attined at the points given by corner points of the feasible region.
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$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
$(i)$ At which point, $Z$ is minimum?
$(ii)$ At which point, $Z$ is maximum ?
$(iii)$ The maximum value of $\mathrm{Z}$ is $\ldots \ldots \ldots$
$(iv)$ The minimum value of $\mathrm{Z}$ is $\ldots \ldots \ldots$
If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:
$\frac{1}{\sqrt{2}}<\text{x}\leq1$
$0\leq\text{x}<\frac{1}{\sqrt{2}}$
$-1\leq\text{x}<\frac{1}{\sqrt{2}}$
$\text{x}>0$
$( S 1): A \cap B =(1, \infty)-N$ and
$( S 2): A \cup B=(1, \infty)$