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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 $P$?

1. $6$
2. $7$
3. $8$
4. $9.5$
Migrated from GO Mechanical 3 years ago by Arjun

### 1 comment

Given :-   $xyz = 192$

$xy = 48$   {putting z=4}

$AM \geq GM \geq HM$

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

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

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

$P \geq 6.928$

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.

by

### 1 comment

You could have simply use the fact that AM >= GM. And this will will give P > 6.xyz and therefore minimum value is 7.

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.

in question x y z are integer so how x= 4*1.73

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

@ 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.

nice :)

you can also write it as an answer.
xyz=192

z=4

xy=192/4

xy=48

p=x+y/2

we all know that least sum of x and y is 14 (x=8 , y=6)

which means p=7

but there comes the problem(p!=7)

in question it is said that x and y are integers

so we can take -1 and -48 whose product is 48 and average is -24.5

and -24.5<7