Example 38 - Prove that radius of cylinder of greatest curved

Example 38 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Let OC = r be the radius of cone & OA = h be height of cone & ∠ OAQ = α be the semi−vertical angle of cone Let OE = x be the radius of cylinder Height of cylinder = OO’ From (1) & (2) ﷐𝑥﷮ℎ − ﷐𝑂﷮′﷯𝑂﷯=﷐𝑟﷮ℎ﷯ ﷐ℎ𝑥﷮𝑟﷯=ℎ−﷐𝑂﷮′﷯𝑂 ﷐𝑂﷮′﷯𝑂=ℎ−﷐ℎ𝑥﷮𝑟﷯ ﷐𝑂﷮′﷯𝑂= ﷐ℎ𝑟 − ℎ𝑥﷮𝑟﷯ ﷐𝑂﷮′﷯𝑂= ﷐ℎ﷐𝑟 − 𝑥﷯﷮𝑟﷯ Now, Curved Surface Area Of cylinder = 2𝜋× Radius of cylinder × Height of cylinder S = 2𝜋 × 𝑥 × ﷐𝑂﷮′﷯𝑂 S = 2𝜋𝑥﷐ℎ﷐𝑟 − 𝑥﷯﷮𝑟﷯ S = ﷐2𝜋ℎ﷮𝑟﷯﷐𝑟𝑥−﷐𝑥﷮2﷯﷯ S = 𝑘﷐𝑟𝑥−﷐𝑥﷮2﷯﷯ We need to minimize S, So, finding S’(x) S’ = ﷐𝑑﷐𝑘﷐𝑟𝑥 − ﷐𝑥﷮2﷯﷯﷯﷮𝑑𝑥﷯ S’ = 𝑘 ﷐𝑑﷐𝑟𝑥 − ﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ S’ = 𝑘﷐𝑟−2𝑥﷯ Putting S’ = 0 0 = 𝑘﷐𝑟−2𝑥﷯ 𝑟−2𝑥 = 0 𝑥 = ﷐𝑟﷮2﷯ Now, Finding S’’(x) at x = ﷐𝑟﷮2﷯ S’’ = ﷐𝑑﷐𝑘﷐𝑟 − 2𝑥﷯﷯﷮𝑑𝑥﷯ S’’ = 𝑘 ﷐𝑑﷐𝑟 − 2𝑥﷯﷮𝑑𝑥﷯ S’’ = 𝑘 ﷐0−2﷯ S’’ = −2𝑘 So, S’’ = −2𝑘 Hence, at 𝑥=﷐𝑟﷮2﷯ ﷐﷐𝑆′′﷮𝑥 = ﷐𝑟﷮2﷯﷯﷯<0 ∴ 𝒙=﷐𝒓﷮𝟐﷯ is maxima of S. Hence, radius of cylinder with greatest curved surface area which can be inscribed in a given cone is half of that cone.