The supremum value of $x^{2}y^{2}(x^{2}+y^{2})$ when $x+y=2n$ for some fixed $n\in \mathbb N $ Let  $S$  be  the  set  of  all  tuples $(x,y)$  such  that $x+y=2n$    for  a  fixed  $n\in \mathbb N$. Then what  is  the  supremum  value of   $x^{2}y^{2}(x^{2}+y^{2})$ $?$

I  substituted $y=2n-x$ $\ $ in $\ $ $x^{2}y^{2}(x^{2}+y^{2})$ and  naming  it  $f(x)$  tried  differentiating  so  at $f'(x)=0$  the  supremum  value . I  hope  that  method  was  correct. But  the  differentiation  was  too tedious and since  this  problem  was  set  in  an  exam  for  $2$  marks  only  I  have  doubt  that  there  is  some  shorter  trick to  get it . Is  there $?$ Please  let  me  know.

Thanks  for  any  help.
$$f(x,y)=x^2y^2\{(x+y)^2-2xy\}=4n^2t^2-2t^3=g(t)$$ where $t=xy$

Use Second derivative test 

and honour $4xy\le(x+y)^2=(2n)^2$
$x_m = n + m, \quad y_m = n - m, \quad m \in \mathbb{N} $,  then $x_m + y_m = 2n$ and $x_m^{2}y_m^{2}(x_m^{2}+y_m^{2})$ goes to infinite when $m \rightarrow \infty $. Therefore there is no supremum value.