Answer
The maximum area of a rectangle inscribed in the ellipse is $2a b$.
Work Step by Step
Using the information in Figure 16, the area of the rectangle is given by $f\left( {x,y} \right) = \left( {2x} \right)\left( {2y} \right) = 4xy$.
Our task is to maximize the function $f\left( {x,y} \right)$ subject to the constraint $g\left( {x,y} \right) = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {4y,4x} \right) = \lambda \left( {\frac{{2x}}{{{a^2}}},\frac{{2y}}{{{b^2}}}} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $4y = \frac{{2\lambda x}}{{{a^2}}}$, ${\ \ \ }$ $4x = \frac{{2\lambda y}}{{{b^2}}}$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
Thus, $\lambda = \frac{{2y{a^2}}}{x} = \frac{{2x{b^2}}}{y}$
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain ${y^2} = \frac{{{b^2}}}{{{a^2}}}{x^2}$.
So, $y = \pm \frac{b}{a}x$.
Substituting it in the constraint gives
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{a^2}}} - 1 = 0$
$\frac{{2{x^2}}}{{{a^2}}} = 1$
So, $x = \pm \frac{a}{{\sqrt 2 }}$.
Using $y = \pm \frac{b}{a}x$ we obtain the critical points: $\left( {\frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right)$, $\left( {\frac{a}{{\sqrt 2 }}, - \frac{b}{{\sqrt 2 }}} \right)$, $\left( { - \frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right)$, $\left( { - \frac{a}{{\sqrt 2 }}, - \frac{b}{{\sqrt 2 }}} \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {\frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right)}&{2ab}\\
{\left( {\frac{a}{{\sqrt 2 }}, - \frac{b}{{\sqrt 2 }}} \right)}&{ - 2ab}\\
{\left( { - \frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right)}&{ - 2ab}\\
{\left( { - \frac{a}{{\sqrt 2 }}, - \frac{b}{{\sqrt 2 }}} \right)}&{2ab}
\end{array}$
From the results in this table, we conclude that the maximum area of a rectangle inscribed in the ellipse is $2a b$.
