find the maximum value of x^3y^3+3xy when x+y=8

• In questions where sum is constant is given and then maximum value has to be found out... then we divide that constant sum in equal parts and put its value in the required equation......
  
  
  x+y=8
  x=8/2
  y=8/2
  put in equation
  
  (4^3*4^3)+3*4*4=4144
• 11 years agoHelpfull: Yes(27) No(0)
• you can use Am>=Gm
  where Am=(x+y)/2 =8/2 =4
  and gm=sqrt(xy)
  4>=sqrt(xy)
  16>=xy
  now, x^3y^3=(16*16)^3 =4096
  3xy = 16*3= 48
  therefore 4096+48 =4144 is the maximum value
  
• 11 years agoHelpfull: Yes(11) No(7)
• x^3y^3=(16*16)^3 ? ? can anyone explain this?
  
• 11 years agoHelpfull: Yes(4) No(0)
• What is the same question asked for a minimum number?? Any chances??
  Thanks :)
• 11 years agoHelpfull: Yes(0) No(0)
• By substituting the values of x=4 and y=4
  
  (4 ^ 3*4 ^ 3)+3*4*4
  (64*64)+3*16
  4096+48
  =4144
  
• 9 years agoHelpfull: Yes(0) No(0)