Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 22

Determine the area of the largest rectangle that can be inscribed in the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} =1$.



Area of the square is $A = 2x(2y) = 4xy$
We can rewrite the equation of the ellipse as...
$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 \quad \Longleftarrow \quad \frac{b^2x^2 + a^2 y^2}{a^2b^2} =1$

Solving for $y$,

$
\begin{equation}
\begin{aligned}
y^2 &= \frac{a^2b^2 - b^2x^2}{a^2} = \frac{b-(a^2-x^2)}{a^2}\\
\\
y &= \sqrt{\frac{b^2}{a^2} (a^2 - x^2)}\\
\\
y &= \frac{b}{a} \sqrt{a^2 - x^2}
\end{aligned}
\end{equation}
$

Substituting the value of $y$ to the equatioon of the area.
$\displaystyle A = 4x \left( \frac{a}{b} \sqrt{a^2 - x^2} \right)$
Taking the derivative of the area, we get...
$\displaystyle A' = \frac{4b}{a} \left[ x \left( \frac{1}{2\sqrt{a^2 - x^2}} \right) (-2x) + (1) \sqrt{a^2 - x^2} \right]$

when $A' = 0$

$
\begin{equation}
\begin{aligned}
0 &= \frac{-x^2}{\sqrt{a^2 - x^2}} + \sqrt{a^2 - x^2}\\
\\
0 &= \frac{-x^2 + a^2 - x^2}{\sqrt{a^2 - x^2}}\\
\\
0 &= -2x^2+a^2\\
\\
2x^2 = a^2
\end{aligned}
\end{equation}
$


Then, the critical number is $\displaystyle x = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$
if $\displaystyle x = \frac{a}{\sqrt{2}}$, then

$
\begin{equation}
\begin{aligned}
y &= \frac{b}{a} \sqrt{a^2 - \left( \frac{a}{\sqrt{2}} \right)^2} = \frac{b}{a} \sqrt{a^2 - \frac{a^2}{2}} = \frac{b}{a} \sqrt{\frac{a^2}{2}}\\
&= \frac{b}{\cancel{a}} \left( \frac{\cancel{a}}{\sqrt{2}} \right) = \frac{b}{\sqrt{2}}
\end{aligned}
\end{equation}
$


Therefore, the largest area that can be described in the ellipse is $\displaystyle A = 4xy = 4 \left( \frac{a}{\sqrt{2}} \right) \left( \frac{b}{\sqrt{2}} \right) = 2ab$