Question-13
The height of a right circular cone of maximum volume that can be enclosed within a hollow sphere of radius \(\displaystyle R\) is:
Invoking symmetry, we have:
It is safe to assume that \(\displaystyle \theta \in ( 0,\pi /2)\). The expression for the volume as a function of \(\theta\) is:
\[ \begin{equation*} V_{\text{cone}} =\frac{1}{3} \pi ( R\cos \theta )^{2}[ R( 1+\sin \theta )] \varpropto \cos^{2} \theta ( 1+\sin \theta ) \end{equation*} \]
We can now rewrite the expression to be optimized as:
\[ \begin{equation*} \left( 1-\sin^{2} \theta \right)( 1+\sin \theta ) =-\sin^{3} \theta -\sin^{2} \theta +\sin \theta +1 \end{equation*} \]
Letting \(\displaystyle y=\sin \theta\) and noting that \(\displaystyle y\in ( 0,1)\), we get:
\[ \begin{equation*} f( y) =-y^{3} -y^{2} +y+1 \end{equation*} \]
\(\displaystyle f^{\prime }( y) =0\Longrightarrow 3y^{2} +2y-1=0\Longrightarrow y=\frac{1}{3} ,-1\). We can ignore \(\displaystyle -1\) as it is extraneous. Since \(\displaystyle f^{\prime \prime} \left(\frac{1}{3}\right) < 0\), we conclude that \(\displaystyle y=\frac{1}{3}\) is a local maximum. Since \(\displaystyle f( 0) =1\), \(\displaystyle f( 1) =0\) and \(\displaystyle f\left(\frac{1}{3}\right) >1\), we conclude that \(\displaystyle \frac{1}{3}\) corresponds to an absolute maximum in \(\displaystyle ( 0,1)\) for \(\displaystyle f\).
Plugging this back in the expression for height, we get \(\displaystyle h=R\left[ 1+\frac{1}{3}\right] =\frac{4}{3} R\).