I have a rectangle inscribed in the parabola defined by

$\displaystyle y=8-x^2$ with one edge on the x-axis ?

Would I assume that edge on the x-axis is x or 2x?

Printable View

- May 3rd 2009, 11:48 AMsk8erboyla2004Rectangle Inscribed in an Area
I have a rectangle inscribed in the parabola defined by

$\displaystyle y=8-x^2$ with one edge on the x-axis ?

Would I assume that edge on the x-axis is x or 2x? - May 3rd 2009, 11:51 AMderfleurer
Your width is 2x, your height is y.

- May 3rd 2009, 11:57 AMsk8erboyla2004
So area = b*h

$\displaystyle 2x(8-x^2)$

and differentiate

$\displaystyle (2x(-2x))+2(8-x^2)$

$\displaystyle -4x^2+16+2x^2$

Solve for 0

$\displaystyle -2x^2+16$

$\displaystyle +-\sqrt{8}$

which one I use? positive or negative - May 3rd 2009, 12:02 PMderfleurer
So i'm guessing you're trying to maximize your area?

$\displaystyle A = 2x(8 - x^2) = 16x - 2x^3$

$\displaystyle A' = 16 - 6x^2$

$\displaystyle 0 = 16 - 6x^2$

$\displaystyle 16 = 6x^2$

$\displaystyle x = \sqrt \frac{16}{6}$

Seeing as we can't have negative dimensions, we use +. - May 3rd 2009, 12:07 PMsk8erboyla2004
Thanks man it just seems I keep getting all sorts of things confused (Headbang)

I get the dimsenions are

$\displaystyle \sqrt{\frac{16}{6}}*\frac{16}{3}$ - May 3rd 2009, 12:17 PMderfleurer
$\displaystyle A = 2xy$

$\displaystyle x = \sqrt {16/6}$

$\displaystyle y = 8 - \sqrt {16/6}^2 = 8 - 8/3 = 16/3$

Looks good to me. Just remember, technically, our base is 2x, not x.