12th Class Mathematics Sample Paper Mathematics Sample Paper-11

  • question_answer
    Find the volume of the largest cylinder that can be inscribed in sphere of radius r.
    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
    \[\Rightarrow \]   \[\pi {{r}^{2}}-\frac{3\pi {{h}^{2}}}{4}=0\Rightarrow h=\frac{2}{\sqrt{3}}r\]
    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
    \[\Rightarrow \]               \[{{R}^{2}}={{r}^{2}}-\frac{1}{4}\times {{\left( \frac{2}{\sqrt{3}}r \right)}^{2}}=\sqrt{\frac{2}{3}}r\]
    \[\therefore \]Maximum volume of the cylinder is given by
    \[{{V}_{\max }}=\pi {{R}^{2}}h=\pi {{\left( \sqrt{\frac{2}{3}}r \right)}^{2}}\left( \frac{2}{\sqrt{3}}r \right)=\frac{4\pi {{r}^{3}}}{3\sqrt{3}}cu\,\,units\]
    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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