Question-22
Find the area of the rectangle with largest perimeter that can be inscribed in a circle of unit radius.
\(2\)
If the angle made by the diagonal of the rectangle with the diameter is \(\displaystyle \theta\), the sides are \(\displaystyle 2\cos \theta\) and \(\displaystyle 2\sin \theta\).
We have to maximize:
\[\begin{equation*} \underset{\theta }{\max} \ 4(\cos \theta +\sin \theta ) \end{equation*}\] subject to:
\[\begin{equation*} 0\leqslant \theta \leqslant \frac{\pi }{2} \end{equation*}\]
Since \(\displaystyle f( \theta ) =\cos \theta +\sin \theta\) is continuous in \(\displaystyle \left[ 0,\frac{\pi }{2}\right]\), it has an absolute maximum and minimum in this interval.
\[\begin{equation*} \begin{aligned} f^{\prime }( \theta ) & =\cos \theta -\sin \theta \end{aligned} ;\ \ \ f^{\prime \prime }( \theta ) =-\sin \theta -\cos \theta \end{equation*}\]
In the given interval, \(\displaystyle f^{\prime }( \theta ) =0\Longrightarrow \theta =\frac{\pi }{4}\). At this point, \(\displaystyle f^{\prime \prime }\left(\frac{\pi }{4}\right) < 0\), which implies that the point is a local maximum. \(\displaystyle f\left(\frac{\pi }{4}\right) =\sqrt{2}\) and this is greater than \(\displaystyle f( 0) =f\left(\frac{\pi }{2}\right) =1\). It follows that \(\displaystyle \frac{\pi }{4}\) corresponds to the point of absolute maximum of \(\displaystyle f\) in this interval. The area of the corresponding rectangle (square) is \(\displaystyle \boxed{2}\).