Question

Find the area of the largest rectangle that can be inscribed in

the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).

Short answer

The area of the largest rectangle is \(2ab\).

Step-by-step solution

Application of Derivative

We can apply the derivative of a function to find the closest point of a curve to a point or the furthest point of a curve to another point. We find these points with the concept of local minimaand local maximarespectively.

We find local minima and maxima of a function by finding critical pointsand the second derivativeof the function.

Critical points are the \(x\) values that satisfy \(f'\left( x \right) = 0\).

The function has local maxima at a point \(x\) if \(f''\left( x \right) < 0\).

The function has a local minimum at a point \(x\) if \(f''\left( x \right) > 0\).

Finding equations related to a rectangle and ellipse

Let the sides of the rectangle be \(2x,2y\) .

Hence the area of the rectangle is \(4xy\) .

Also, from the equation of ellipse we have,

\(\begin{aligned}{c}y = b\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} \\y = \frac{b}{a}\sqrt {{a^2} - {x^2}} \end{aligned}\)

Hence \(A\left( x \right) = 4\frac{b}{a}x\sqrt {{a^2} - {x^2}} \) .

Finding the Local Maxima of the area function

Differentiating we get,

\(\begin{aligned}{c}A'\left( x \right) = 4\frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} - \frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}} \right)\\A'\left( x \right) = 4\frac{b}{a}\left( {\frac{{{a^2} - {x^2} - {x^2}}}{{\sqrt {{a^2} - {x^2}} }}} \right)\\A'\left( x \right) = \frac{{4b}}{{a\sqrt {{a^2} - {x^2}} }}\left( {{a^2} - 2{x^2}} \right)\end{aligned}\)

So, the critical point will be,

\(\begin{aligned}{c}A'\left( x \right) = 0\\4\frac{b}{a}\left( {{a^2} - 2{x^2}} \right) = 0\\\left( {{a^2} - 2{x^2}} \right) = 0\\2{x^2} = {a^2}\\x = \frac{1}{{\sqrt 2 }}a\end{aligned}\)

Hence the critical point is \(x = \frac{1}{{\sqrt 2 }}a\).

The second derivative will be \(A''\left( x \right) = - 16\frac{b}{a}x < 0\).

Hence the maximum value of the function \(A\) will be at the point \(x = \frac{1}{{\sqrt 2 }}a\).

Hence the other side will be,

\(\begin{aligned}{l}y = \frac{b}{a}\sqrt {{a^2} - \frac{1}{2}{a^2}} \\y = \frac{1}{{\sqrt 2 }}b\end{aligned}\)

Thus, the area of the largest rectangle will be,

\(\begin{aligned}{c}A = 4\frac{1}{{\sqrt 2 }}a\frac{1}{{\sqrt 2 }}b\\ \Rightarrow A = 2ab\end{aligned}\)

Thus, the area of the largest rectangle is \(2ab\).