How Do You Find The Maximum Area Of A Rectangle In An Eclipse?

**AlphaRock** · March 26th 2009, 06:35 PM

(deleted)

**mollymcf2009** · March 26th 2009, 07:14 PM

Originally Posted by AlphaRock

Can somebody show me step by step how to get the maximum area of a rectangle in an eclipse? Because I'm burnt out.

$(x^2)/9 + (y^2)/4 = 1$

Thanks!

Area of a rectangle = $xy$

Equation of the ellipse = $\frac{x^2}{9} + \frac{y^2}{4} = 1$

First solve your ellipse equation for x.

$\frac{4x^2 + 9y^2}{36} = 1<br /> <br />$
$4x^2 + 9y^2 = 36$

....

$x = ^{+}_{-} \sqrt{\frac{36-9y^2}{4}}$

Plug this into your area equation:

$A(y) = (\sqrt{\frac{36-9y^2}{4}})y<br />$
Take your derivative and put it = to zero. Solve. Then just plug in your value for y into the ellipse equation that you solved for x above and this will give you your dimensions of your rectangle with the largest area. Then just use your values to find an exact area.

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