Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Example 38 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
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 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 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 of 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 (Where k = 2𝜋ℎ/𝑟 is a constant) 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.
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Example 51
Chapter 6 Class 12 Application of Derivatives
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.