Example 38 - Prove that radius of cylinder of greatest curved

Example 38 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 38 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 38 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 38 - Chapter 6 Class 12 Application of Derivatives - Part 5

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

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 cone. Let OC = r be the radius of cone & OA = h be height of cone & ∠ OAQ = α be the semi-vertical angle of cone And, Let OE = x be the radius of cylinder Height of cylinder = OO’ Since cone is given, radius (r) and height (h) of cone are constant And, radius (x) and height (OO’) of cylinder is variable Finding Height OO’ in terms of h, r and x From (1) & (2) 𝒙/(𝒉 − 𝑶^′ 𝑶)=𝒓/𝒉 ℎ𝑥/𝑟=ℎ−𝑂^′ 𝑂 𝑂^′ 𝑂=ℎ−ℎ𝑥/𝑟 𝑂^′ 𝑂= (ℎ𝑟 − ℎ𝑥)/𝑟 In Δ AO’Q tan α = (𝑂^′ 𝑄)/𝐴𝑂′ tan α = 𝑂𝐸/(𝑂𝐴 − 𝑂^′ 𝑂) tan 𝜶 = 𝑥/(ℎ − 𝑂^′ 𝑂) In ∆𝑨𝑶𝑪 tan α = 𝑂𝐶/𝑂𝐴 tan 𝜶 = 𝑟/ℎ 𝑶^′ 𝑶= 𝒉(𝒓 − 𝒙)/𝒓 Now, Curved Surface Area of Cylinder = 2𝜋 × Radius × Height 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 (Where k = 2𝜋ℎ/𝑟 is a constant) Finding S’(x) S’ (x) = 𝑑(𝑘(𝑟𝑥 − 𝑥^2 ))/𝑑𝑥 S’ (x) = 𝑘 𝑑(𝑟𝑥 − 𝑥^2 )/𝑑𝑥 S’ (x) = 𝑘(𝑟−2𝑥) Putting S’ = 0 0 = 𝑘(𝑟−2𝑥) 𝑟−2𝑥 = 0 𝒙 = 𝒓/𝟐 Finding S’’(x) at x = 𝒓/𝟐 S’’ (x) = 𝑑(𝑘(𝑟 − 2𝑥))/𝑑𝑥 S’’ (x) = 𝑘 𝑑(𝑟 − 2𝑥)/𝑑𝑥 S’’ = 𝑘 (0−2) S’’ = −2𝑘 Therefore, S’’ (x) < 0 for 𝑥=𝑟/2 Thus, 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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.