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Misc 17 - Show that height of cylinder of maximum volume - Miscellaneous

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  1. Class 12
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Misc 17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is ﷐2𝑅﷮﷐﷮3﷯﷯ . Also find the maximum volume. Let R be the radius of sphere Let h be the height & 𝑥 be the diameter of cylinder In ∆ 𝐴𝐵𝐶 Using Pythagoras theorem ﷐﷐𝐶𝐵﷯﷮2﷯+﷐﷐𝐴𝐵﷯﷮2﷯=﷐﷐𝐴𝐶﷯﷮2﷯ h2 + ﷐﷐𝑥﷯﷮2﷯=﷐﷐𝑅+𝑅﷯﷮2﷯ h2 + 𝑥2 =﷐﷐2𝑅﷯﷮2﷯ h2 + 𝑥2 = 4R2 𝑥2 = 4R2 – h2 We need to find maximum volume of cylinder Let V be the volume of cylinder V = π ﷐﷐𝑟𝑎𝑑𝑖𝑢𝑠﷯﷮2﷯ × ﷐ℎ𝑒𝑖𝑔ℎ𝑡﷯ V = π ﷐﷐﷐𝑥﷮2﷯﷯﷮2﷯ ×ℎ V = π × ﷐﷐𝑥﷮2﷯﷮4﷯ ×ℎ V = π ﷐﷐4﷐𝑅﷮2﷯ − ﷐ℎ﷮2﷯﷯﷮4﷯ ×ℎ V = ﷐4﷐𝑅﷮2﷯𝜋ℎ﷮4﷯−﷐𝜋﷐ℎ﷮3﷯﷮4﷯ V = πhR2 – ﷐𝜋﷐ℎ﷮3﷯﷮4﷯ Different w.r.t ℎ ﷐𝑑𝑣﷮𝑑ℎ﷯=﷐𝑑﷐𝜋ℎ﷐𝑅﷮2﷯ − 𝜋﷐ℎ﷮3﷯/4﷯﷮𝑑ℎ﷯ ﷐𝑑𝑣﷮𝑑ℎ﷯= π R2 ﷐𝑑﷐ℎ﷯﷮𝑑ℎ﷯−﷐𝜋﷮4﷯﷐𝑑﷐﷐ℎ﷮3﷯﷯﷮𝑑ℎ﷯ ﷐𝑑𝑣﷮𝑑ℎ﷯= π R2 – ﷐𝜋﷮4﷯ ﷐3﷐ℎ﷮2﷯﷯ ﷐𝑑𝑣﷮𝑑ℎ﷯= π R2 – ﷐3𝜋﷮4﷯ h2 Putting ﷐𝑑𝑣﷮𝑑ℎ﷯=0 π R2 – ﷐3﷮4﷯ 𝜋 ﷐ℎ﷮2﷯=0 ﷐3﷮4﷯𝜋﷐ℎ﷮2﷯=𝜋﷐𝑅﷮2﷯ h2 = ﷐𝜋﷐𝑅﷮2﷯﷮﷐3﷮4﷯𝜋﷯ h2 = ﷐4﷐𝑅﷮2﷯﷮3﷯ h =﷐﷮﷐4﷐𝑅﷮2﷯﷮3﷯﷯ h = ﷐2𝑅﷮﷐﷮3﷯﷯ Finding ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐ℎ﷮2﷯﷯ ﷐𝑑𝑣﷮𝑑ℎ﷯=𝜋﷐𝑅﷮2﷯−﷐3﷮4 ﷯ 𝜋 ﷐ℎ﷮2﷯ Different w.r.t h ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐ℎ﷮2﷯﷯= ﷐𝑑﷐𝜋﷐𝑅﷮2﷯ − ﷐3﷮4﷯ 𝜋﷐ℎ﷮2﷯﷯﷮𝑑ℎ﷯ ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐ℎ﷮2﷯﷯= 0 – ﷐3𝜋﷮4﷯ ×2ℎ ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐ℎ﷮2﷯﷯= ﷐−3𝜋ℎ﷮4﷯ Putting h = ﷐2𝑅﷮﷐﷮3﷯﷯ ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐ℎ﷮2﷯﷯ = ﷐−3𝜋﷮2﷯﷐﷐2𝑅﷮﷐﷮3﷯﷯﷯ = –﷐﷮3﷯𝜋𝑅 < 0 ∴ h = ﷐2𝑅﷮﷐﷮3﷯﷯ is point of maxima So, volume is maximum when h = ﷐2𝑅﷮﷐﷮3﷯﷯ From (1) 𝑥2 = 4R2 – h2 𝑥2 = 4R2 – ﷐﷐﷐2𝑅﷮﷐﷮3﷯﷯﷯﷮2﷯ 𝑥2 = 4R2 – ﷐4﷐𝑅﷮2﷯﷮3﷯ 𝑥2 = ﷐12﷐𝑅﷮2﷯ − 4﷐𝑅﷮2﷯﷮3﷯ 𝑥2 = ﷐8﷮3 ﷯﷐ 𝑅﷮2﷯ Maximum value of volume is V = π ﷐﷐﷐𝑥﷮2﷯﷯﷮2﷯ℎ V = ﷐𝜋﷮4﷯﷐𝑥﷮2﷯ℎ Putting value of 𝑥2 & h V = ﷐𝜋﷮4﷯ × ﷐8﷐𝑅﷮2﷯﷮3﷯ × ﷐2𝑅﷮﷐﷮3﷯﷯ V = ﷐ 16𝜋﷐𝑅﷮3﷯﷮12﷐﷮3﷯﷯ V = ﷐𝟒𝝅﷐𝑹﷮𝟑﷯﷮𝟑﷐﷮𝟑﷯﷯ cubic unit

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