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If x,y,z are integers such that x+y+z=12. Find maximum value of xyz+xy +yz+zx

If x,y,z are integers such that x+y+z=12. Find maximum value of xyz+xy +yz+zx 

Grade:9

2 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
(x+y+z)^2 = 144     {squaring both sides}
       Also using identity we know
       (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)
       144 = x^2 + y^2 + z^2 + 2(xy+yz+zx)
      ½  (144 – (x^2 + y^2 + z^2)) = xy+yz+zx
 For xy+yz+zx to be maximum, x^2 + y^2 + z^2 = 0
  • x=y=z=0
  • x,y,z = 0
 
therefore xyz=0 and value of xy+yz+zx = 72
 
max value of given expression = 72+0
                                                  = 72
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Pritid Mallik
22 Points
one year ago
x+y+z = 12 – (1)
We know that A.M is always greater than or equal to G.M.
43xyz
=) 64xyz – (2)
Now, squaring both sides of eq. (1), we get,
x²+y²+z²+2(xy+yz+xz) = 144
xy+yz+xz = 1/2(144-x²-y²-z²)
                 = 72-1/2(x²+y²+z²) – (3)
Now, to find the maximum value of xy+yz+zx, we need to substitute the smallest value of x²+y²+z² (which is 48, where x=y=z=4) in eq.(3)
∴ RHS of eq.(3) becomes 72-1/2(48) = 72-24 = 48
∴ From eq.(2) and eq.(3), the maximum value of xy+yz+zx+xyz becomes 64+48 =112

Cheers!!
Regards,
Pritid (9th grade
Sri Chaitanya Techno School)

 

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