# Ex 6.5,23 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

Ex 6.5, 23 Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.Cone of largest volume inscribed in the sphere of radius R Let OC = x Radius of cone = BC Height of cone = h = OC + OA Finding OC Δ BOC is a right angled triangle Using Pythagoras theorem in ∆BOC 〖𝑂𝐵〗^2=〖𝐵𝐶〗^2+〖𝑂𝐶〗^2 R2 =〖𝐵𝐶〗^2+𝑥^2 BC2 = 𝑅^2 – x2 BC = √(𝑅^2−𝑥^2 ) Thus, Radius of cone = BC = √(𝑅^2−𝑥^2 ) Height of cone = OC + OA = R + x We need to show Maximum volume of cone = 8/27 × Volume of sphere = 8/27 × 4/3 𝜋𝑅^3 = 32/81 𝜋𝑅^3 Let V be the volume of a cone We know that Volume of a cone = 1/3 𝜋(𝑟𝑎𝑑𝑖𝑢𝑠 )^2 (ℎ𝑒𝑖𝑔ℎ𝑡 ) V = 1/3 𝜋(√(𝑅^2−𝑥^2 ))^2 (𝑅+𝑥) V = 1/3 𝜋(𝑅^2−𝑥^2 )(𝑅+𝑥) V = 1/3 𝜋(𝑅^2 (𝑅+𝑥)−𝑥^2 (𝑅+𝑥)) V = 1/3 𝜋(𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 ) Diff w.r.t 𝒙 𝑑𝑉/𝑑𝑥=𝑑/𝑑𝑥 [1/3 𝜋(𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 )] 𝑑𝑉/𝑑𝑥=𝜋/3 [𝑑/𝑑𝑥 (𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 )] 𝑑𝑉/𝑑𝑥=𝜋/3 (0+𝑅^2.1−𝑅 ×2𝑥−3𝑥^2 ) 𝑑𝑉/𝑑𝑥=𝜋/3 (𝑅^2−2𝑅𝑥−3𝑥^2 ) Putting 𝒅𝑽/𝒅𝒙=𝟎 1/3 𝜋(𝑅^2−2𝑅𝑥−3𝑥^2 )=0 𝑅^2−2𝑅𝑥−3𝑥^2=0 𝑅^2−3𝑅𝑥+𝑅𝑥−3𝑥^2=0 R(𝑅−3𝑥)+𝑥(𝑅−3𝑥)=0 (𝑅+𝑥)(𝑅−3𝑥)=0 So, x = –R & 𝑥=𝑅/3 Since x cannot be negative 𝑥=𝑅/3 Finding (𝒅^𝟐 𝒗)/(𝒅𝒙^𝟐 ) 𝑑𝑣/𝑑𝑥=1/3 𝜋(𝑅^2−2𝑅𝑥−3𝑥^2 ) (𝑑^2 𝑣)/(𝑑𝑥^2 )=𝜋/3 𝑑/𝑑𝑥 (𝑅^2−2𝑅𝑥−3𝑥^2 ) (𝑑^2 𝑣)/(𝑑𝑥^2 )=𝜋/3 (0−2𝑅−6𝑥) (𝑑^2 𝑣)/(𝑑𝑥^2 )=(−𝜋)/3 (2𝑅+6𝑥) Putting 𝒙=𝑹/𝟑 〖(𝑑^2 𝑣)/(𝑑𝑥^2 )│〗_(𝑥 = 𝑅/3) =(−𝜋)/3 (2𝑅+6(𝑅/3)) =(−𝜋)/3 (2𝑅+2𝑅) =(−𝜋)/3 (4𝑅) =(−4𝜋𝑅)/3 < 0 Thus (𝑑^2 𝑣)/(𝑑𝑥^2 )<0 when 𝑥=𝑅/3 ∴ Volume is Maximum when 𝑥=𝑅/3 Finding maximum volume From (1) Volume of cone = 1/3 𝜋(𝑅^2−𝑥^2 )(𝑅+𝑥) Putting 𝑥 = 𝑅/3 = 1/3 𝜋(𝑅^2−(𝑅/3)^2 )(𝑅+𝑅/3) = 1/3 𝜋(𝑅^2−𝑅^2/9)((3𝑅 + 𝑅)/3) = 1/3 𝜋((9𝑅^2− 𝑅^2)/9)(4𝑅/3) = 𝟑𝟐/𝟖𝟏 𝝅𝑹^𝟑 Hence proved

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Chapter 6 Class 12 Application of Derivatives

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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.