Question

This has to do inscribing. find the area of the largest rectangle that can fit into this ellipse: x^2+4y^2=4

Answer 1

So, looking at it. First we need a goal. The largest rectangle will have the largest Area. So we are trying to find the Biggest amount of area.

Answer 2

Hmm I think it might be easier to NOT work in the first quadrant.
This is my thought.Labeling the sides of our rectangle gives us:\[\Large\bf\sf A\quad=\quad (2x)(2y)\quad=\quad 4xy\]And of course we have our constraint,\[\Large\bf\sf x^2+4y^2=4\qquad\to\qquad x=2\sqrt{1-y^2}\]

Answer 3

\[\Large \bf\sf A=4y\left(2\sqrt{1-y^2}\right)\]

Answer 4

\[\Large\bf\sf A'=8\sqrt{1-y^2}+8y\frac{-2y}{2\sqrt{1-y^2}}\]Looking for critical points, we set the derivative function equal to zero and we'll need to find a common denominator:\[\Large\bf\sf 0=\frac{1-y^2-y^2}{\sqrt{1-y^2}}\]Multiply through by the denominator:\[\Large\bf\sf 0=1-2y^2\]Giving us a critical point of:\[\Large\bf\sf y=\frac{\sqrt2}{2}\]

Answer 5

Bravo! I totally got stumped while I was trying to figure it out.

Answer 6

so how do I find the area

Answer 7

So we should determine if this y critical point corresponds to a max or a min. It's going to be a max, but your teacher would probably want you to check.

Answer 8

\[\Large\bf\sf A'(1/4)=\frac{1-2\left(\frac{1}{4}\right)^2}{\sqrt{1-y^2}}\]Gives us a positive number. So the function is increasing on the left of our critical point.