Max/Min problem 
Find the maximum area of a rectangle DACB where C and B are two points
  on the graph of $y=\dfrac{8}{1+x^2}$ and A and D are the two corresponding points on
  the  x axis.

I've been at this for 3 hours now and keep getting the wrong answer, the answer at the back of the book is 8 and the answer I keep getting is 4.
I noticed while I was typing this up that B is not on the graph (you can see in the image below).

I took $y=\dfrac{8}{1+x^2}$ as the height of the rectangle, between points B and D or points C and A. So I substituted $y=\dfrac{8}{1+x^2}$ into the area forumla $A = xy$ getting:
$$A =(x)(\dfrac{8}{1+x^2})$$
$$A =\dfrac{8x}{1+x^2}$$
I diffrentiated that and put it equal to zero to solve for x or the width of the rectangle and received 1 and -1. I know this is the wrong answer as B is not on the graph as I mentioned earlier...I know my area formula is where I'm going wrong, am I to multiply $x$ by 2 at the start before diffrentiating? so it will be $$A =\dfrac{16x}{1+x^2}$$ part of me believes that wouldn't work anyway..I can't my head around what the x value should be as it's confusing because the rectangle is on both 1st and 2nd quadrants...If I can get hint at least that would be great.
 A: You've already realized the mistake. The width should be $2x$ so that $B$ will be on the graph, symmetric to $C$. 
Everything stays the same after that. You will get the same $x$ value but the area is doubled, which will be the correct answer $8$.
A: In order that $B$ and $C$ both lie on the graphics and $ABCD$ is a rectangle, we need $BC\parallel AD$, hence $B$ and $C$ have the same $y$-coordinate, so they are symmetric with respect to the $y$-axis. Assuming $B=\left(x,\frac{8}{1+x^2}\right)$ with $x\geq 0$, we have:
$$ [ABCD] = 8\cdot\frac{2 x}{1+x^2}\leq 8$$
with equality attained at $x=1$.
A: let $A(x/2;0),B(x/2,f(x/2))$ and $D(-x/2,0)$ then the area is given by $$A=xf(x/2)=\frac{32x}{x^2+4}$$
A: The are of the rectangle is its height times its base. Its height, as you noted, is $$\frac{8}{1+x^2}$$. It's base runs from the point $x$ to $-x$, a distance of $2x$. So your thought that the area is
$$M=(2x)\frac{8}{1+x^2}=\frac{16x}{1+x^2}$$ is correct.
Differentiating with respect to $x$ by the quotient rule we get,
$$\frac{dM}{dx}=16\frac{1-x^2}{(1+x^2)^2}.$$
Setting this equal to 0, we see that the maximum of $M$ occurs at $\pm1$, both of which correspond to $|M|=8$.
