• # question_answer Find the volume of the largest cylinder that can be inscribed in sphere of radius r. OR A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of its semicircular ends is $\pi :(\pi +2).$

 Given, r is the radius of sphere. Let R be the radius, h be the height of cylinder and V be the volume of the cylinder. Then,                $V=\pi {{R}^{2}}h$                  ?(i) In right angled $\Delta \,OAC,$we have ${{r}^{2}}={{R}^{2}}+{{\left( \frac{h}{2} \right)}^{2}}$ $\Rightarrow$               ${{R}^{2}}={{r}^{2}}-\frac{{{h}^{2}}}{4}$ $\therefore$                  $V=\pi {{r}^{2}}h-\frac{\pi {{h}^{3}}}{4}$ $\therefore$                  $\frac{dV}{dh}=\pi {{r}^{2}}-\frac{3\pi {{h}^{2}}}{4}$ Now,                 $\frac{{{d}^{2}}V}{d{{h}^{2}}}=0-\frac{6\pi h}{4}=\frac{-\,3\pi h}{2}$ For maximum or minimum value of V, put $\frac{dV}{dh}=0$ $\Rightarrow$   $\pi {{r}^{2}}-\frac{3\pi {{h}^{2}}}{4}=0\Rightarrow h=\frac{2}{\sqrt{3}}r$ Now, ${{\left( \frac{{{d}^{2}}V}{d{{h}^{2}}} \right)}_{h=\frac{2}{\sqrt{3}}r}}=\frac{-\,3\pi }{2}\times \frac{2}{\sqrt{3}}r=-\,\sqrt{3}\pi r<0$ Thus, V is maximum when$h=\frac{2}{\sqrt{3}}r$. $\therefore$R is calculated as ${{R}^{2}}={{r}^{2}}-\frac{{{h}^{2}}}{4}$ $\Rightarrow$               ${{R}^{2}}={{r}^{2}}-\frac{1}{4}\times {{\left( \frac{2}{\sqrt{3}}r \right)}^{2}}=\sqrt{\frac{2}{3}}r$ $\therefore$Maximum volume of the cylinder is given by ${{V}_{\max }}=\pi {{R}^{2}}h=\pi {{\left( \sqrt{\frac{2}{3}}r \right)}^{2}}\left( \frac{2}{\sqrt{3}}r \right)=\frac{4\pi {{r}^{3}}}{3\sqrt{3}}cu\,\,units$ OR Let r be radius of semicircular end and h be the height of the half cylinder. Volume of half cylinder,$V=\frac{1}{2}\pi {{r}^{2}}h;$    ?(i) Total surface area, $S=\pi rh+\pi {{r}^{2}}+2rh$ $S=(\pi +2)\,r\cdot \frac{2V}{\pi {{r}^{2}}}+\pi {{r}^{2}}=\frac{2V\,(\pi +2)}{\pi r}+\pi {{r}^{2}}$ [from Eq. (i)] $\frac{dS}{dr}=\frac{-\,2V\,(\pi +2)}{\pi {{r}^{2}}}+2\pi r$                               ?(ii) For minimum or minimum surface area, put $\frac{dS}{dr}=0$ $\Rightarrow$               $\frac{2V\,(\pi +2)}{\pi {{r}^{2}}}=2\pi r$       ?(iii) $\frac{{{d}^{2}}S}{d{{r}^{2}}}=\frac{4V\,(\pi +2)}{\pi {{r}^{3}}}+2\pi >0$for ${{r}^{3}}=\frac{2V\,(\pi +2)}{2{{\pi }^{2}}}$ [from Eq. (iii)] $\therefore$S is minimum for $\frac{V\,(\pi +2)}{\pi {{r}^{2}}}=\pi r$       [from Eq. (iii)] $\Rightarrow$               $\frac{\frac{1}{2}\pi {{r}^{2}}h\,(\pi +2)}{\pi {{r}^{2}}}=\pi r$    [from Eq. (i)] $\Rightarrow$               $\frac{h}{2r}=\frac{\pi }{\pi +2}\Rightarrow h:2r=\pi :\pi +2$ Hence proved.

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