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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The radius of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is [AMU 1999]

    A) \[\frac{2}{3}R\]

    B) \[\sqrt{\frac{2}{3}}R\]

    C) \[\frac{3}{4}R\]

    D) \[\sqrt{\frac{3}{4}}R\]

    Correct Answer: B

    Solution :

    • If r be the radius and h the height, then from the figure,   
    • \[{{r}^{2}}+{{\left( \frac{h}{2} \right)}^{2}}={{R}^{2}}\]Þ \[{{h}^{2}}=4({{R}^{2}}-{{r}^{2}})\]           
    • Now, \[V=\pi {{r}^{2}}h\] = \[2\pi {{r}^{2}}\sqrt{{{R}^{2}}-{{r}^{2}}}\]           
    • \ \[\frac{dV}{dr}=4\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}+2\pi {{r}^{2}}.\frac{1}{2}\frac{(-2r)}{\sqrt{{{R}^{2}}-{{r}^{2}}}}\]           
    • For max. or min., \[\frac{dV}{dr}=0\]           
    • Þ  \[4\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}=\frac{2\pi {{r}^{3}}}{\sqrt{{{R}^{2}}-{{r}^{2}}}}\] Þ \[2({{R}^{2}}-{{r}^{2}})={{r}^{2}}\]           
    • Þ  \[2{{R}^{2}}=3{{r}^{2}}\] Þ \[r=\sqrt{\frac{2}{3}}R\]  Þ \[\frac{{{d}^{2}}V}{d{{r}^{2}}}=-ve\].           
    • Hence V is max., when \[r=\sqrt{\frac{2}{3}}R\].


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