# Thread: Rectangle Inscribed in an Area

1. ## Rectangle 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?

3. 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

4. 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 +.

5. Thanks man it just seems I keep getting all sorts of things confused

I get the dimsenions are

$\displaystyle \sqrt{\frac{16}{6}}*\frac{16}{3}$

6. $\displaystyle A = 2xy$

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

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

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