How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$ 
let $x,y\in (0,1)$, and such
  $$(1+x)(1+y)=81(1-x)(1-y)$$
  Prove 
  $$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask  $\dfrac{9}{16}$ is best constant?

PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)(10-8y)=\dfrac{1}{64\cdot 81}(8+8x)(8+8y)(10-8x)(10-8y)$$
since
$$(8+8x)(10-8x)\le\dfrac{18^2}{4}=81,(8+8y)(10-8y)\le 81$$
if and only if $$8+8x=10-8x,8+8y=10-8y\Longrightarrow x=y=\dfrac{1}{8}$$
but this not such
$$(1+x)(1+y)=81(1-x)(1-y)$$
so How find this maximum?

Let $P=xy$, $S=x+y$.  Rewrite both the condition and the other formula in terms of $P$ and $S$.  Use the condition to eliminate $P$, and maximize as a function of $S$.  Also check the edges of the square.
