if x+y+z=21 then find the maximum value of (x+2) (y+2) (z+2)

• out of many possibilities, 7+7+7=21 gives the maximum value
  
  9*9*9=729
• 8 years agoHelpfull: Yes(5) No(0)
• for 3 nos. arithmetic mean> geometric mean.
  consider nos. x+2,y+2,z+2
  their am is (a+b+c)/3
  (x+y+z+6)/3=9
  g.m. is [(x+2)(y+2)(z+2)]^1/3
  am>gm
  taking cube both side
  729 >(x+2)(y+2)(z+2)
  HENCE max. value is 729
• 8 years agoHelpfull: Yes(3) No(0)
• x,y and z will have different values so I guess x=8,y=7,z=6, as 6+7+8=21
  so answer will be= 10*9*8=720
• 8 years agoHelpfull: Yes(0) No(1)
• The possible solutions to get 21 are =20,1,0.
  .
  .
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  9,9,3--(9+2)(9+2)(5)----max
• 8 years agoHelpfull: Yes(0) No(1)
• 729
  9*9*9=729
  
• 8 years agoHelpfull: Yes(0) No(0)
• remainder=3
  2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256
  from above observe last digits 2,4,8,6,2,4,6,8,............... here 2^35 th last digit is 8 ,so 8/5=3.
• 8 years agoHelpfull: Yes(0) No(0)