1. ## Largest Rectangle

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?

I wasn't at school for the lesson so I don't know where to begin. Any help would be nice. Thanks.

2. Originally Posted by lancelot854
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?

I wasn't at school for the lesson so I don't know where to begin. Any help would be nice. Thanks.
$12-x^2=(\sqrt{12}-x)(\sqrt{12}+x)$

This crosses the x-axis at $\pm\sqrt{12}$

It is symmetrical about the y-axis so we can maximise the half of the rectangle that lies on the positive side of the x-axis.

The entire rectangle area A is $2x\left(12-x^2\right)=24x-2x^3$

This is a maximum when the tangent to the graph is horizontal, hence differentiate A and set the derivative to zero to find x.

3. You know that the length of this rectangle is $2x$ (simply from drawing it), and the height of this rectangle is $y$, but for consistency's sake, rewrite it as height $= 12 - x^2$. So, the area of our rectangle is,
$A(x) = 2x(12 - x^2) = 24x - 2x^3$. You should use the first derivative test to find out when this maximizes.

4. Thanks for the help guys, however, I just don't understand where the $2x$ in this problem comes from. How could I find this? Thanks again.

5. The picture of a rectangle tangent to a parabola is here, (generically):
http://www.ms.uky.edu/~carl/ma123/ko...chap14e115.gif
We want the horizontal length of this rectangle. Well, the rectangle starts (horizontally at some point on the graph $(-x, y)$, and ends at the point $(x, y)$. So the length is simply $x - (-x) = 2x$.

6. Ok, it all makes sense now, thanks a lot.