i think 7 is correct answer . x and y is 8,6

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+3 votes

The product of three integers $X$, $Y$ and $Z$ is $192$. $Z$ is equal to $4$ and $P$ is equal to the average of $X$ and $Y$. What is the minimum possible value of $P$?

- $6$
- $7$
- $8$
- $9.5$

Migrated from GO Mechanical 1 year ago by Arjun

+2 votes

**Given**

$X*Y*Z = 192$,

$Z=4,$

$P = \left ( \frac{X+Y}{2} \right )$

**Solution**

$X*Y*Z = 192$

$\Rightarrow$ $X*Y*4 = 192$

$\Rightarrow$ $X*Y = 48$

Now $48$ should be broken into product of 2 numbers such that they are close to each other because if numbers are close to each other then their average will be minimum..

For eg:- $48$ can be $4*12$, $1*48$, $6*8$, $16*3$ but here we will select $6*8$ because $6$ and $8$ are closest to each other with a difference of $2$.

$\Rightarrow$ $X*Y = 6*8$

$X=6, Y=8$

$P = \left ( \frac{X+Y}{2} \right ) = \left ( \frac{8+6}{2} \right )= 7$

$\therefore$ Option B. $7$ is the correct answer.

+2 votes

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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