A rectangle is inscribed in the ellipse

\[ \frac {x^2}{3}+\frac {y^2}{5}=1 \]
so that two of its edges are parallel to the \(x\)-axis, and the other two are parallel to the \(y\)-axis. Find the dimensions of the rectangle with the largest area.
Area of the rectangle \(A=2x\cdot 2y\). We will express \(A\) as a function of, say, \(x\). Note that \(x\) and \(y\) satisfy the equation of the ellipse, so
\[ y^2=5- \frac {5x^2}{3} \]
Since in our figure \(y\ge 0\), we have
\[ y=\sqrt {5- \frac {5x^2}{3}} \]
Now \(A(x)=2x\cdot 2\sqrt {5- \frac {5x^2}{3}}\).

The domain of \(A=\left [0,\answer {\sqrt {3}}\right ]\).

We have to find the critical points of \(A\).

\(A'(x)=4\cdot \sqrt {5- \frac {5x^2}{3}}+\frac {4x}{2\sqrt {5- \frac {5x^2}{3}}}\cdot \left (-\frac {10}{3}x\right )\). If we simplify \(A'(x)\), we get

\(A'(x)=4\cdot \sqrt {5- \frac {5x^2}{3}}-\frac {20x^2}{3\sqrt {5- \frac {5x^2}{3}}}\),

and if we simplify again, we get

\(A'(x)=\frac {60-40x^2}{3\sqrt {5- \frac {5x^2}{3}}}\)

The function \(A\) has the only critical point at \(x=\sqrt {\frac {\answer {3}}{2}}\).

Since the domain of \(A\) is a closed interval \(\left [0,\answer {\sqrt {3}}\right ]\), we have to evaluate the function \(A\) at the end points and at the critical point:

\[ A(0)=0 \]
\[ A(\sqrt {3})=0 \]
\[ A\left (\sqrt {\frac {\answer {3}}{2}}\right )=4\sqrt {\frac {\answer {3}}{2}}\sqrt {5- \frac {5}{2}}=2\sqrt {15}. \]
So, the maximum is attained at the critical point.
\begin{align*} \text {base}&= \answer {\sqrt {6}}\\ \text {height} &= \answer {\sqrt {10}} \end{align*}