Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$ Additional Info
$x < 1, y<2, x+y>0$
I tried doing AM-GM (since all of the terms are positive.
$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$
this way we can cancel the x and y's 
thus the maximum value is 
$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$
but the answer is said to be 1(?) where $x =0,$ and $y = 1$
Can someone tell me what I did wrong.

In your AM-GM the equality occurs for
$$2-2x=2-y=2x+2y=\frac{1}{4},$$ which is impossible. 
I think, it's better to make the following.
By AM-GM
$$(1-x)(2-y)^2(x+y)=4(1-x)\left(1-\frac{y}{2}\right)^2(x+y)\leq$$
$$\leq4\left(\frac{1-x+2\left(1-\frac{y}{2}\right)+x+y}{4}\right)^4=\frac{81}{64}.$$
The equality occurs for $$1-x=1-\frac{y}{2}=x+y$$ or
$$(x,y)=\left(\frac{1}{4},\frac{1}{2}\right),$$ which says that we got a maximal value.
