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

• ANSWER: 4144
  x^3 * y^3 + 3x*y = (xy)^3 + 3xy
  Now x+y=8 and we have to maximize xy
  x=1 y=7 gives xy=7
  x=2 y=6 gives xy=12
  x=3 y=5 gives xy=15
  x=4 y=4 gives xy=16 (MAXIMUM)
  Hence x=4 and y=4
  Solve 16(256+3)= 4144
• 11 years agoHelpfull: Yes(33) No(3)
• i think when both x and y are 4,3x*y is max=48 then ans will be 4^36+48...bt how to calculate this??:-(anyone post the ans...
• 11 years agoHelpfull: Yes(2) No(1)
• i think 4096 maybe the answer....
• 11 years agoHelpfull: Yes(2) No(5)
• here x=4 and y=4 because x+y=8
  given expression is X^3*Y^3+3*X*Y
  so 4^3*4^3+3*4*4
  =64*64+48
  =4144 this is tha answer
• 9 years agoHelpfull: Yes(2) No(0)
• plz..tell me how the result would be come out
• 11 years agoHelpfull: Yes(1) No(1)
• x^3 * y^3 + 3x*y = (xy)^3 + 3xy
  Now x+y=8 and we have to maximize xy
  x=1 y=7 gives xy=7
  x=2 y=6 gives xy=12
  x=3 y=5 gives xy=15
  x=4 y=4 gives xy=16 (MAXIMUM)
  Hence x=4 and y=4
  Solve 16(256+3)= 4144
  
• 9 years agoHelpfull: Yes(1) No(1)