# Rectangle Inscribed in an Area

• May 3rd 2009, 11:48 AM
sk8erboyla2004
Rectangle Inscribed in an Area
I have a rectangle inscribed in the parabola defined by

$y=8-x^2$ 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 AM
derfleurer
• May 3rd 2009, 11:57 AM
sk8erboyla2004
So area = b*h

$2x(8-x^2)$

and differentiate

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

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

Solve for 0

$-2x^2+16$

$+-\sqrt{8}$

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

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

$A' = 16 - 6x^2$

$0 = 16 - 6x^2$

$16 = 6x^2$

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

Seeing as we can't have negative dimensions, we use +.
• May 3rd 2009, 12:07 PM
sk8erboyla2004
Thanks man it just seems I keep getting all sorts of things confused (Headbang)

I get the dimsenions are

$\sqrt{\frac{16}{6}}*\frac{16}{3}$
• May 3rd 2009, 12:17 PM
derfleurer
$A = 2xy$

$x = \sqrt {16/6}$

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

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