Find the area of the greatest rectangle that can be inscribed in an ellipse.

Question

Q. Find the area of the greatest rectangle that can be inscribed in the ellipse.

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\tag1$$

Before anyone says, I have already seen the previously posted queries. Link: Find the area of largest rectangle that can be inscribed in an ellipse

During my initial attempts, I tried to calculate the area for the rectangle in the first quadrant and multiplied it by 4.

Thus $$\text{Area} = 4(a-x)(b-y).\tag2$$

where a and b are the major and minor axes(interchangeable) which are subtracted from variables x and y, respectively.

I then substituted y from (1) in (2)

$$\text{Area}(A) = (4)(b - \frac{b}{a}\sqrt{a^2-x^2})(x-a)$$

I then took the derivative of $A$ and equated it to $0$ giving $x = a/\sqrt2$ and $y = b/\sqrt2$.

I then took the second order derivative of $A$ which gives a negative value. Thus, A is maximum at $x =a/\sqrt2$.

The issue occurs when I substituted the previously found $x$ and $y$ values in (2). I get $A = 2ab(\sqrt2 - 1)$.

I just need clarification whether this approach is viable or not. If yes, where did I go wrong.

Answer 1

According to the definition of $x$ and $y$ used in the equation of the ellipse, the area of the rectangle is $$A=\color{red}{4xy}.\tag1$$ This means: $$ A=4x b\sqrt{1-\frac{x^2}{a^2}} $$ Equating to $0$ the derivative of $A$ over $x$ one obtains: $$ x=\frac a{\sqrt2},\quad y=\frac b{\sqrt2}. $$

Substituting this in $(1)$ one obtains: $$ A=2ab. $$

Your error is highlighted above using the red color. If you change the meaning of the variables, you have to change all equations involving the variables according to the new meaning, not just one of them.

Answer 2

Your strange decision to re-define $x$ and $y$ is what causes your problem. Using these coordinates, (1) has to be $$\frac{(a-x)^2}{a^2}+\frac{(b-y)^2}{b^2}=1$$ and this is what you must use when you substitute for $y$.