Yathiraj   9 years AGO

The e possible values will be 2,2,4 (problem doesn't say the number can't repeat)
To get the minimum possible value of x*y^2*z^2, x should be assigned with biggest value of above.
Hence,
x=4
y=2
z=2
With above values, x*y^2*z^2 = 4*4*4 = 64

Option 3.

Like? Yes (9) | No (2)  

DEEPAK   9 years AGO

The values of x y z can be:
1. 2,2,4
2.2,4,2
3.4,2,2

The minimum value can only be possible for (3).
Thus, x*y^2*z^2=4*4*4=64

Like? Yes (2) | No   

praveen kumar   9 years AGO

the value of x y z are 4,2,2
so ans is 64

Like? Yes (2) | No (1)  

Samyak   9 years AGO

x+y+z =8, xy+yz+xz =20
So solving the equation you can have value 2,2,4
So, minimum value in x*y2*z2 will be 64. So option 3.

Like? Yes (4) | No (3)  

aman aditya   9 years AGO

X=2,y=2,z=4
So value is 64

Like? Yes (1) | No   

Prithwijit Ghosh   9 years AGO

X,Y,Z ARE NON ZERO NO
SO,WE KNOW 2+2+4=8
2*2+2*4+4*2=20
SO ANSWER WILL BE 4*2^2*2^2=64

Like? Yes (2) | No (1)  

varun bansal   9 years AGO

let x=4,y=2,z=2.
then 4+2+2=8 {first condition satisfied}
then (4*2)+(2*2)+(2*4)=20 {2nd cond. satisfied}
To find min of :
x*y^2*z^2
: 4*(2^2)*(2^2)
= 64.

Like? Yes (1) | No   

PRASHANT KUMAR   9 years AGO

X+Y+Z=8;
X^2+Y^2+Z^2=24;
and xy+yz+xz=20;
so only possible values are (4,2,2) or (2,4,2) or (2,2,4)
but the only possible value to give the minimum is 4,2,2.

Like? Yes (1) | No   

DEEPAK SINGHAL   9 years AGO

th combination is 4,2,2,

4*4*4=64

Like? Yes (2) | No (1)  

durgesh singh   9 years AGO

x=4, y=2=z

Like? Yes (1) | No   

arush gupta   9 years AGO

ans 64

Like? Yes (1) | No   

Josepheena Joseph   9 years AGO

The possible solution for x,y,z is 2,2,4
minimum value is 4*2^2*2^2=64

Like? Yes (1) | No   

ganesh babu   9 years AGO

x=4, y=2, z=2
x=2, y=4, z=2
x=2, y=2, z=4
min (4*2^2*2^2, 2*4^2*2^2, 2*2^2*4^2)
min(64, 128, 128)
ans:64

Like? Yes (1) | No   

Pragati Rode   9 years AGO

from this equation, we get x=4,y=2 and z=2
minimum possible value is 64

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

X=4, Y=Z= 2
So answer is 64

Like? Yes (1) | No   

Madhumoy Gope   9 years AGO

x+y+z=8
xy+yz+zx=20
So,the possible values of x,y,z are 4,2,2
So,minimum possible value of x*y^2*z^2=4*2^2*2^2=64
So,option (3) is correct

Like? Yes (1) | No   

Supriya Mondal   9 years AGO

From the question we get,
x=4,y=2,z=2
So, minimum value of x*y^2*z^2 will be=4*2^2*2^2
=64
So, option 3)64 is the right answer

Like? Yes (1) | No   

Abhik Das   9 years AGO

The numbers are=4,2,2
means x=4,y=2,z=2
so the min value will be=4*4*4=64

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

value of x,y,z=4,2,2

x*y+Y*z+z*x=20
2*2+2*4+4*2=20

x*y^2+z^2=(2*2)^2*4^2
4*4*4
16*4=64

Like? Yes (1) | No   

Vaibhav Srivastava   9 years AGO

trial and error

Like? Yes (1) | No   

b.subbarao   9 years AGO

x+y+z=8
let the values of x,y,z are be 4,2&2 respectively
then xy+yz+zx=4*2+2*2+2*4=20
x*y^2*z^2=4*2^2*2^2=64

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

X,Y,Z = 2,2,4
x*y^2*z^2 = 4*2^2*2^2 = 4*4*4 = 64

Like? Yes (1) | No   

vivek raj   9 years AGO

x+y+z=8 ......(1)
xy+yz+zx=20 .........(2)

eq 2-1x
x^2-8x+(20-yz)=0

x^2-8x+16=0

from comparing we get

20-yz=16
yz=4

putting value of yz in eq 2
xy+xz=16

from eq 1
y+z=8-x

x(y+z)=16
x(8-x)=16
x^2-8x+16=0
x=4
yz=4

so ans x*y^2*z^2=64

Like? Yes (1) | No (1)  

RAKESH   9 years AGO

only soln set is (4,2,2)

so, x*y^2*z^2 = 64 or 128
minm value of x*y^2*z^2 will be 64 .

Like? Yes (1) | No (1)  

RAHUL KUMAR   9 years AGO

X=4, Y=2, Z=2
then X*Y^2*Z^2=64

Like? Yes (1) | No (1)  

Jaspalsingh Parmar   9 years AGO

Option 3 is corrrect

Like? Yes (1) | No (1)  

KUMAR HARSH   9 years AGO

x+y+z=8 ---(1)
xy+yz+xz=20
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
x^2+y^2+z^2=24---(2)
from 1 and 2
we get x=4,y=2,z=2
x*y^2*z^2 =4*2^2*2^2=64
so ans is 64

Like? Yes (1) | No (1)  

PAWAN KUMAR   9 years AGO

From above equation, we have(for minimum value) :
x=4, y=2, z=2
So, x*y^2*z^2 = 4*4*4 = 64 Ans.

Like? Yes (1) | No (1)  

Devendra Marghade   9 years AGO

Possible combinations of (x,y,z) with x+y+z = 8 and
xy+yz+zx = 20 are (2,2,4) , (2,4,2), (4,2,2)
And minimum possible value of x*y^2*z^2=4*(2^2)*(2^2)=64
Ans=Option (3)

Like? Yes (1) | No (1)  

sravani   9 years AGO

4,2,2 is the ordertriplet which satisfies both conditions so, min possible value of x*y^2*z^2=64

Like? Yes (1) | No (1)  

asif khan   9 years AGO

x=4 y=2 z=2

Like? Yes (1) | No (1)  

Jay parmar   9 years AGO

option 3 is correct

Like? Yes (7) | No (8)