Maximum Volume of Cylinder in a Cone

Given a cone of fixed height  
\[H\]
  and base radius  
\[R\]
, what are the radius of the base and the height of the cylinder that give the maximum volume?

From the diagram,  
\[\frac{H-h}{}= \frac{r}{R} \rightarrow r= \frac{R(H-h)}{H}\]
.
The volume of a cylinder is given by
\[V= \pi r^2h= \pi (\frac{R(H-h)}{H})^2h=\frac{\pi R^2}{H^2}(H^2h-2Hh^2+h^3)\]

\[\frac{dV}{dh}= \frac{\pi R^2}{H^2}(H^2-4Hh+3h^2) \]

The volume is maximum when  
\[\frac{dV}{dh}=0 \rightarrow H^2-4Hh+3h^2=0 \rightarrow (H-3h)(H-h)=0\]
.
Then  
\[h=0 \rightarrow V=0\]
  or  
\[h= \frac{H}{3}\]
  and  
\[r=R \frac{H-h}{H}=R \frac{H-H/3}{H}= \frac{2}{3}R\]
.
The maximum volume is  
\[V= \pi r^2h= \pi (\frac{2R}{3})^2 \frac{H}{3}= \frac{4 \pi R^2H}{27}\]
.

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