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?

Hint.   Let $X=\frac{1+x}{1-x},\ Y=\frac{1+y}{1-y}$. We have $XY=81$ and $1\le X,\ 1\le Y$. Since $$(1-x)(10-8x)=\left(1-\frac{X-1}{X+1}\right)\left(10-8\frac{X-1}{X+1}\right)=\frac{4(X+9)}{(X+1)^2},$$ what we need to show is 
$$\frac{(X+9)}{(X+1)^2}\cdot\frac{(Y+9)}{(Y+1)^2}\le \frac9{256}$$
and we would like to know when the equality holds.
Continuing from Hint.  Let $k=X+Y$. 
$$\frac{(X+9)}{(X+1)^2}\cdot\frac{(Y+9)}{(Y+1)^2}=\frac{162+9k}{(82+k)^2}.$$
Noting $18\le k\le 82$ and by differentiating the above by $k$, we find that $\frac{162+9k}{(82+k)^2}$ attains its maximum at $k=46$. The maximum value is
$$\frac{162+9\cdot 46}{(82+46)^2}=\frac{576}{128^2}=\frac{9}{256},$$
as required.
(If you don't want to use differentiation, you could instead use the AM-GM inequality:
\begin{align*}
\frac{162+9k}{(82+k)^2}&=\frac{9(k+18)}{(k+18)^2+128(k+18)+64^2}\\
&=\frac{9}{(k+18)+\frac{64^2}{k+18}+128}\\
&\le \frac{9}{2\sqrt{64^2}+128},
\end{align*}
where the last equality holds when $k+18=64$.)
