Calculus "Word Problem" Here is the problem:

A rectangle has its base on the $x$-axis and its upper two vertices on the parabola $y=12-x^2$ What is the largest area the rectangle can have, and what are its dimensions? 

Well, I don't really know where to start. My initial idea was to find inflection points because I figured that is where the vertices would be, but there are no inflection points because it is a parabola.
Then I though about finding where the derivative and the parabola cross, found it, but I don't know how that will help me. 
I really don't know where to start. 
Any help is appreciated, thanks. 
 A: Have a look at following picture. Does this help? Can you get (using this picture) area of the rectangle?
You should get $A(x)=2x(12-x^2)$ as mentioned in André's comment.
After getting this expression for $A(x)$, can you  continue with the solution? What will be $A'(x)$ and where is it equal to zero?
You have $A(x)=24x-2x^3$, which means
$$A'(x)=24-6x^2.$$
Solving $A'(x)=0$ gives $x=2$ and $A(x)=2\cdot2\cdot(12-2^2)=2\cdot2\cdot8=32$ as you correctly stated in the comment.
(There is also solution $x=-2$ but that one is not interesting for us - we are looking for $x\ge 0$ since we denote by $x$ the point to the right from $0$.)

A: The function $y=12-x^2$ is symmetrical to the $y$-axis, so in order for the upper corners of the rectangle to be on the curve, they must be at x-values that are opposites.
The length of the bottom of the rectangle is $2x$.
The height is $12-x^2$.
So the area of the rectangle at a value $x$ is:
$$A(x) = 2x(12-x^2) = 24x-2x^3.$$
$$A'(x) = 24-6x^2.$$
We set this equal to zero to determine the $x$-value which makes $A(x)$ maximum.
$$24-6x^2=0 \Rightarrow 4=x^2\Rightarrow x=\pm 2.$$
We take $x=2$.
So the area is maximized when the bottom corners are at $(2,0)$ and $(-2,0)$.
