# Maximum possible area

• Mar 31st 2008, 07:20 PM
pennydooodle
Maximum possible area
My first mistake was taking a Calculus class online... my second mistake was trying to learn Calculus from the textbook...

Here is the problem that I'm having a hard time understanding:

What is the maximum possible area of a rectangle inscribed into the ellipse
[(x^2) + (4y^2) = 4] with the sides of the rectangle parallel to the coordinate axis?
• Mar 31st 2008, 08:12 PM
TheEmptySet
Quote:

Originally Posted by pennydooodle
My first mistake was taking a Calculus class online... my second mistake was trying to learn Calculus from the textbook...

Here is the problem that I'm having a hard time understanding:

What is the maximum possible area of a rectangle inscribed into the ellipse
[(x^2) + (4y^2) = 4] with the sides of the rectangle parallel to the coordinate axis?

Attachment 5640

from the diagram we see that the area of the rectangle in the first quadrant is $\displaystyle A=xy$ so the total area is $\displaystyle A=4xy$

solving for x in the ellipse we get $\displaystyle x=2\sqrt{1-y^2}$

subbing into the Area formula we get

$\displaystyle A=8y \sqrt{1-y^2}$

taking the derivative

$\displaystyle \frac{dA}{dy}=8\left( \sqrt{1-y^2}+\frac{y}{2\sqrt{1-y^2}}(-2y)\right)=8\left( \frac{(1-y^2)-y^2}{\sqrt{1-y^2}}\right)$

setting the numerator equal to zero we get

$\displaystyle 1-2y^2=0 \iff y=\frac{\sqrt{2}}{2}$

subbing y into the equation for x we get $\displaystyle x=2\sqrt{1-\left( \frac{\sqrt{2}}{2}\right)^2}=2\sqrt{1-1/2}=\sqrt{2}$

So The area of the whole rectangle is $\displaystyle A=4xy=4\sqrt{2}\frac{\sqrt{2}}{2}=4$