Application of Derivative to determine from the Maximum volume the dimensions of given solid geometry

Exercise : (Application of Derivative )

Find the Radius and height of the right circular cylinder of largest volume that can be inscribed in a right circular cone with radius 6 cm and height 8 cm .

Solution: Let $r$ be the radius of Cylinder inscribed in the cone and

$h$ be the respective height for that cylinder .

Observe that $\left( DE \right)//\left( CB \right)$ due to the cylinder fact

Hence $C\hat{E}D={{90}^{\circ }}$ thus $\vartriangle ABC\approx \vartriangle EDA$ by A.A

       $(A)\,\,\,A\hat{D}E=A\hat{C}B={{90}^{\circ }}$

       $(A)\,\,C\hat{A}B=D\hat{A}E$ ( Common angle )

So the ratio of similarity will be as follows :

$\left. \frac{ADE}{ACB} \right)=\frac{AD}{AC}=\frac{AE}{AB}=\frac{DE}{CB}\Leftrightarrow \frac{8-h}{8}=\frac{AE}{AB}=\frac{r}{6}$

So $\frac{8-h}{8}=\frac{r}{6}\Leftrightarrow 6\left( 8-h \right)=8r\Leftrightarrow 48-6h=8r$

Note that the volume for Cylinder is $V=\pi {{r}^{2}}h$

So $V=\pi {{r}^{2}}\left( 8-\frac{8}{6}r \right)$ with $0\le r\le 6$

So we get $f\left( r \right)=\pi {{r}^{2}}\left( 8-\frac{8}{6}r \right)=8\pi {{r}^{2}}-\frac{8\pi {{r}^{3}}}{6}$

To obtain the max we need to find the derivative of $f$ w.r.t $r$

Hence $f'\left( r \right)=\frac{df}{dr}=16\pi r-\frac{24\pi {{r}^{2}}}{6}=16\pi r-4\pi {{r}^{2}}$

$f'\left( r \right)=0\Leftrightarrow 16\pi r-4\pi {{r}^{2}}=0\Leftrightarrow 4\pi r\left( 4-r \right)=0$ So $r=0\,\,or\,\,r=4$

As result the max of $f$ occur either at either $f'\left( r \right)=0$ or at the end point of$\left[ 0,6 \right]$

$V\left( 0 \right)=0\,\,\And \,\,V\left( 6 \right)=8\pi {{\left( 6 \right)}^{2}}-\frac{8\pi {{\left( 6 \right)}^{3}}}{6}=0$

so the max volume occur at $r=4\,\,cm\,\,\,\,\And \,\,\,\,h=\frac{48-8\times 4}{6}=\frac{8}{3}\,\,cm$