It is given that :- $X*Y*Z= 192$ , $Z=4$. So, $X*Y=48$

Now, $P=\frac{X+Y}{2}=\frac{X+\frac{48}{X}}{2}=\frac{1}{2}\left ( X+\frac{48}{X} \right )$

Now, to find the minimum value of $P,$ Differentiate $P$ with respect to $X$

$\frac{\mathrm{d} P}{\mathrm{d} X} = \frac{1}{2}\left ( 1 -\frac{48}{X^{2}} \right )$

Now, $\frac{\mathrm{d} P}{\mathrm{d} X} = 0$

So, $X= \pm 4\sqrt{3}$

Now, at $X= +4\sqrt{3}$, $\frac{\mathrm{d}^{2}P }{\mathrm{d} X^{2}} > 0$

So, $P$ will be minimum at $X= +4\sqrt{3}$

So, Minimum value of $P$ will be $\frac{1}{2}\left ( 4\sqrt{3} + \frac{48}{4\sqrt{3}} \right ) = 2\sqrt{3} + \frac{6}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} =6.928….$(An Irrational value)

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