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?
 A: Another way is to use the symmetry to reduce a variable.  With $p = x+y$ and $q = xy$, the constraint is $40q = 41p-40$ and you want to maximise
$$(1-x)(1-y)(10-8x)(10-8y) = (1-p+q)(100-80p+64q)$$
Now you can substitute for one variable from the constraint and get to maximise $\frac9{50}p(5-2p)$.
Now that is maximised when $2p(5-2p)$ is maximised, which is when $2p=5-2p \implies p = \frac54$ and the maximum is indeed $\dfrac9{16}$.

P.S. It may be noted that $p = \frac54 \implies q = \frac9{32} \implies \{x, y\} = \{\frac{5-\sqrt7}8, \frac{5+\sqrt7}8 \}$ when the maximum is achieved.
A: We have $$\frac{81}1=\frac{1+xy+(x+y)}{1+xy-(x+y)}$$
Applying Componendo and dividendo,
$$\frac{1+xy}{x+y}=\frac{41}{40}$$
Let $$\frac{1+xy}{41}=\frac{x+y}{40}=u$$
$$\implies(1-x)(1-y)(10-8x)(10-8y)$$
$$=\{1+xy-(x+y)\}\{36+64(1+xy)-80(x+y)\}$$
$$=(41u-40u)\cdot36(1-16u)=\frac{36}{64}\left[1-\left(8u-1\right)^2\right]\le\frac{36}{64}$$
The equality occurs if $8u-1=0$
A: The standard method is Lagrange Multipliers.  Let
$$f(x,y,\lambda)=(1-x)(1-y)(10-8x)(10-8y)+\lambda\left[(1+x)(1+y)-81(1-x)(1-y)\right]$$
Then solve the three equations $\partial f/\partial x=0$,$\partial f/\partial y=0$,$\partial f/\partial \lambda=0$
That covers the inside of the square.  Then check along the sides of the square, where $x=0$, $x=1$, $y=0$ or $y=1$.
A: 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.
A: 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 $k+18=64$.)
