Rectangle Inscribed in an Area

**sk8erboyla2004** · May 3rd 2009, 11:48 AM

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?

**derfleurer** · May 3rd 2009, 11:51 AM

Your width is 2x, your height is y.

**sk8erboyla2004** · May 3rd 2009, 11:57 AM

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

**derfleurer** · May 3rd 2009, 12:02 PM

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 +.