Question

Find the maximum and minimum
values of the equation x3+y3=3xy​

Answer 1

Answer:

Let  f(x,y)=x3y3+3xy=xy((xy)2+3) .

Now,  f(0,0)=0  and  f(1,1)=4 . Therefore  maxf(x,y)≥0 .

Note that,  f(x,y)≥0  for  x,y≥0 .

And,  f(−x,−y)=f(x,y)  and  f(−x,y)=f(x,−y)=−f(x,y) .

Therefore, we can restrict our attention to just  x,y≥0 . Now, note that  f(x,y)  is maximized when  xy  is maximized. Therefore, our problem now is,

maxxy  such that  x+y=8  and  x,y≥0 .

Let  g(x)=x(8−x) . Then,  g′(x)=8−2x  and  g′′(x)=−2 . Therefore,  xy  is maximum at  x=4,y=4 . Hence,  f(x,y)  is maximum at  xy=16 , which gives  maxf(x,y)=163+3×16=4144 .

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