Question

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height \(h\) and semi vertical angle \(α\) is one-third that of the cone and the greatest volume of cylinder is \(\frac{4}{27}πh^3 \,tan^2α\)

Answer 1

The given right circular cone of fixed height \((h)\) and semi-vertical angle \((α)\) can be drawn as: 
Here, a cylinder of radius \(R\) and height \(H\) is inscribed in the cone. 
Then, \(∴GAO=α, OG=r,OA=h,OE=R\), and \(CE=H. \)
We have,\(r=h\,tanα\) 
Now, since \(∆AOG\) is similar to \(∆CEG\), we have:
\(\frac{AO}{OG}=\frac{CE}{EG}\)
\(⇒\frac{h}{r}=\frac{H}{r-R}\,\,\,\,   [EG=OG-OE]\)
\(⇒H=\frac{h}{r}(r-R)\)
\(=\frac{h}{h\,tanα}(h\,tanα-R)\)
\(=\frac{1}{tanα}(h\,tanα-R)\)
Now, the volume \((V)\) of the cylinder is given by,
\(V=πR^2H=\frac{πR^2}{tanα}(h\,tanα-R)\)
\(=πR^2h-\frac{πR^3}{tanα}\)
\(∴\frac{dV}{dR}=2πRh-\frac{3πR^2}{tanα}\)
Now \(\frac{dV}{dR}=0\)
\(⇒2πRh=\frac{3πR^2}{tanα}\)
\(⇒2h\,tanα=3R\)
\(⇒R=\frac{2h}{3}tanα\)
Now,\(\frac{d^2V}{dR^2}=2πh-\frac{6πR}{tanα}\)
And for \(R=\frac{2h}{3tanα}\),we have
\(\frac{d^2V}{dR^2}=2πh-\frac{6π}{tanα}(\frac{2h}{3}tanα)\)
\(=2πh-4πh=-2πh<0\)
∴By second derivative test, the volume of the cylinder is the greatest when
\(R=\frac{2h}{3}tanα\)
When \(R=\frac{2h}{3}tanα,H=\frac{1}{tanα}(htanα-\frac{2h}{3}tanα)\)
\(=\frac{1}{tanα}(\frac{h\,tanα}{3})=\frac{h}{3}\)
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. 
Now, the maximum volume of the cylinder can be obtained as:
\(π(\frac{2h}{3}tanα)^2(\frac{h}{3})=π(\frac{4h^2}{9}tan^2α)(\frac{h}{3})\)
\(=\frac{4}{27}πh^3tan^2α\)
Hence, the given result is proved