what is the maximum value of x^3+y^3+3x*y . when x+y = 8 ??

• As it is not specified that value of x and y are greater than 0 or they can be equal to zero
  If we take y as 0 then maximum value will be 512 .
  and if both x and y are greater than 0 than maximum value will be 365.
• 8 years agoHelpfull: Yes(7) No(0)
• Answer is 365.
  Now x+y=8
  so 1+7=8 , xy =7
  2+6=8, xy =12
  3+5=8, xy = 15
  4+4=8, xy =16
  If xy =16 then we get the value is 176.
  If xy=7..then x=1 and y=7
  so x^3+y^3+3 x*y =1^3+7^3+3*1*7 =365..
  others two value 197 , 260
  so the maximum value 365.
  
• 8 years agoHelpfull: Yes(5) No(0)
• let x=y=4 and put the values of x and y in the expression i. e
  (4^3)+(4^3)+3*4*4=176
• 8 years agoHelpfull: Yes(1) No(3)
• (x+y)^3= x^3 + y^3 + 3xy(x+y)
  so,
  (x+y)^3= x^3 + y^3 + 3xy*8 [x+y=8]
  
  (x+y)^3= x^3 + y^3 + 24xy
  
  (x+y)^3 - 21xy= x^3 + y^3 + 3xy
  
  8^3 -21xy=x^3 + y^3 + 3xy
  
  512 -21xy=x^3 + y^3 + 3xy
  
  so the value of x^3 + y^3 + 3xy will be maximum for the minimum value of xy and that is when xy=7.
  hence,512- (21*7) =365.
• 8 years agoHelpfull: Yes(1) No(0)
• given x+y=8
  for max the value of x and y must be 4,4
  so ans will be 176
• 8 years agoHelpfull: Yes(0) No(0)