Question

Find the area of the largest rectangle that can be inscribed in the ellipse \({{{x^2}} \mathord{\left/

{\vphantom {{{x^2}} {{a^2}}}} \right.

\kern-\nulldelimiterspace} {{a^2}}} + {{{y^2}} \mathord{\left/

{\vphantom {{{y^2}} {{b^2}}}} \right.

\kern-\nulldelimiterspace} {{b^2}}} = 1\).

Short answer

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

Step-by-step solution

To sketch the figure

Since the rectangle is inscribed in the given ellipse, the rectangle must be symmetric about the\(y - \)axis.

The sketch is shown in the figure below:

 To find the area of the rectangle

Let\(x > 0\)be the\(x - \)coordinate and\(y > 0\)be the\(y - \)coordinate of the vertex of the rectangle in the first quadrant.

Therefore, the area of the rectangle is,

\(\begin{array}{l}A = \left( {2x} \right) \cdot \left( {2y\,\,} \right)\,\\ \Rightarrow A = 4xy \cdots \cdots \left( 1 \right)\end{array}\).

We have,

\(\begin{array}{l}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\\ \Rightarrow \frac{{{y^2}}}{{{b^2}}} = 1 - \frac{{{x^2}}}{{{a^2}}}\\ \Rightarrow y = b\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} \end{array}\)

Thus, equation \(\left( 1 \right)\) becomes,

\(A\left( x \right) = 4x\left( {b\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} } \right)\)

In order to find the area of the largest rectangle, we need the derivative of the area function.

Therefore,

\(\begin{array}{l}A'\left( x \right) = 4b\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} + 4bx\left( {\frac{1}{{2\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} }}} \right) \cdot \frac{{ - 2x}}{{{a^2}}}\\ \Rightarrow A'\left( x \right) = \frac{{4b\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right) - 4b\frac{{{x^2}}}{{{a^2}}}}}{{\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} }}\end{array}\)

\(\begin{array}{l} \Rightarrow A'\left( x \right) = \frac{{\frac{{4{a^2}b - 8b{x^2}}}{{{a^2}}}}}{{\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2}}}} }}\\ \Rightarrow A'\left( x \right) = \frac{{4b\left( {{a^2} - 2{x^2}} \right)}}{{a\sqrt {{a^2} - {x^2}} }}\end{array}\)

The derivative is defined whenever\({a^2} - {x^2} > 0 \Rightarrow - a < x < a\).

Now, we find the zeros.

\(\begin{array}{l}A'\left( x \right) = 0\\ \Rightarrow \frac{{4b\left( {{a^2} - 2{x^2}} \right)}}{{a\sqrt {{a^2} - {x^2}} }} = 0\\ \Rightarrow 4b\left( {{a^2} - 2{x^2}} \right) = 0\\ \Rightarrow {a^2} - 2{x^2} = 0\\ \Rightarrow {a^2} = 2{x^2}\\ \Rightarrow x = \frac{a}{{\sqrt 2 }}\end{array}\)

Thus, the only critical number for\(A\)is,\(x = \frac{a}{{\sqrt 2 }}\).

Here,\(A'\left( x \right) < 0,\,\,\forall x > \frac{a}{{\sqrt 2 }}\)\(\)and\(A'\left( x \right) > 0,\,\,\forall x < \frac{a}{{\sqrt 2 }}\).

Hence,\(A\)has an absolute maximum at\(x = \frac{a}{{\sqrt 2 }}\).

Thus, the greatest area of the inscribed rectangle is,

\(\begin{array}{l}A\left( {\frac{a}{{\sqrt 2 }}} \right) = 4\left( {\frac{a}{{\sqrt 2 }}} \right)\left( {b\sqrt {1 - \frac{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}}{{{a^2}}}} } \right)\\ \Rightarrow A\left( {\frac{a}{{\sqrt 2 }}} \right) = \frac{{4ab}}{{\sqrt 2 }}\sqrt {1 - \frac{{{a^2}}}{{2{a^2}}}} \\ \Rightarrow A\left( {\frac{a}{{\sqrt 2 }}} \right) = 4b\frac{a}{2}\\ \Rightarrow A\left( {\frac{a}{{\sqrt 2 }}} \right) = 2ab\end{array}\)