How to find the maximum area of a rectangle that can be inscribed in an ellipse $4x^2+y^2=4$

Question

We choose the point A(x,y) to lie on the ellipse and one of the corners of the rectangle. Let $f$ be the area of the inscribed rectangle. Show that it can be written as: $$f(x)=8x\sqrt{1-x^2}, 0<x<1$$ then find the point A on the ellipse that gives the biggest area

Answer 1

Let $(a \cos t, b \sin t)$ be in first quadrant. The area of the rectangle in this quadrant is $ab \sin t \cos t$. The area of the big rectangle is $$A=4ab \sin t \cos t= 2ab \sin 2t \implies A_{max}=2ab, ~\text{when}~ 2t =\pi/2$$

Answer 2

There is a non-calculus solution.

Convince yourself that sides of optimum rectangle will be parallel to axes of ellipse (or the coordinate axes) due to symmetry of ellipse. As pointed out in comments, its vertices will be

$$ {(x,y), (-x,y), (-x,-y), (x,-y)} $$

which gives sides $2x$, $2y$ and area $S = 4xy$.

Now consider,

$$ (2x-y)^2 \ge 0 $$ $$ \Rightarrow 4x^2 + y^2 - 4xy \ge 0 $$ $$ \Rightarrow S \le 4x^2 + y^2 = 4 $$

The equality occurs at $2x-y=0$ from which one can find point $A$.

Answer 3

Let $(x,y)$ be a vertex of the rectangle.

Thus, by AM-GM $$S=2|x|\cdot2|y|=4|xy|\leq2\cdot\frac{(2|x|)^2+|y|^2}{2}=4.$$ The equality occurs for $2|x|=|y|,$ which says that we got a maximal value.