__Given__ :- $xyz = 192$

$xy = 48$ **{putting z=4}**

$AM \geq GM \geq HM$

$(x+y) /2 \geq \sqrt{xy}$

$(x+y) /2 \geq \sqrt{48}$

$(x+y) /2 \geq 6.928$

$P \geq 6.928$

P=7 is answer.

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6 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 3 years ago by Arjun

5 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)

No options are matching.

**For this question, Marks were given to all.**

0

@ ankitgupta.1729 , can’t we use the concept of AM and GM here? i.e. for any two non negative real numbers, their Arithmetic mean is always greater than or equal to their Geometric mean (AM GM reference)

In this case, Arithmetic mean of X and Y is P. From value of z , we get their geometric mean as $\sqrt{48}$. So according to the inequality, the minimum value of P is $6.928$ , which is closrst to option B.

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