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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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 827 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 𝑂𝐵2= 𝐵𝐶2+ 𝑂𝐶2 R2 = 𝐵𝐶2+ 𝑥2 BC2 = 𝑅2 – x2 BC = 𝑅2− 𝑥2 Thus, Radius of cone = BC = 𝑅2− 𝑥2 Height of cone = OC + OA = R + x We need to show Maximum volume of cone = 827 × Volume of sphere = 827 × 43𝜋 𝑟3 = 3281𝜋 𝑟3 Let V be the volume of a cone We know that Volume of a cone = 13𝜋 𝑟𝑎𝑑𝑖𝑢𝑠 2 ℎ𝑒𝑖𝑔ℎ𝑡 V = 13𝜋 𝑅2− 𝑥22 𝑅+𝑥 V = 13𝜋 𝑅2− 𝑥2 𝑅+𝑥 V = 13𝜋 𝑅2 𝑅+𝑥− 𝑥2 𝑅+𝑥 V = 13𝜋 𝑅3 + 𝑅2𝑥− 𝑥2𝑅− 𝑥3 Diff w.r.t 𝑥 𝑑𝑉𝑑𝑥= 𝑑𝑑𝑥 13𝜋 𝑅3 + 𝑅2𝑥− 𝑥2𝑅− 𝑥3 𝑑𝑉𝑑𝑥= 𝜋3 𝑑𝑑𝑥 𝑅3 + 𝑅2𝑥− 𝑥2𝑅− 𝑥3 𝑑𝑉𝑑𝑥= 𝜋3 0+ 𝑅2.1−𝑅 ×2𝑥−3 𝑥2 𝑑𝑉𝑑𝑥= 𝜋3 𝑅2−2𝑅𝑥−3 𝑥2 Putting 𝑑𝑉𝑑𝑥=0 13𝜋 𝑅2−2𝑅𝑥−3 𝑥2=0 𝑅2−2𝑅𝑥−3 𝑥2=0 𝑅2−3𝑅𝑥+𝑅𝑥−3 𝑥2=0 R 𝑅−3𝑥+𝑥 𝑅−3𝑥=0 𝑅+𝑥 𝑅−3𝑥=0 So, x = –R & 𝑥= 𝑅3 Since x cannot be negative 𝑥= 𝑅3 Finding 𝑑2𝑣𝑑 𝑥2 𝑑𝑣𝑑𝑥= 13𝜋 𝑅2−2𝑅𝑥−3 𝑥2 𝑑2𝑣𝑑 𝑥2= 𝜋3 𝑑𝑑𝑥 𝑅2−2𝑅𝑥−3 𝑥2 𝑑2𝑣𝑑 𝑥2= 𝜋3 0−2𝑅−6𝑥 𝑑2𝑣𝑑 𝑥2= −𝜋3 2𝑅+6𝑥 Putting 𝑥= 𝑅3 𝑑2𝑣𝑑 𝑥2│𝑥= 𝑅3 = −𝜋3 2𝑅+6 𝑅3 = −𝜋3 2𝑅+2𝑅 = −𝜋3 4𝑅 = −4𝜋𝑅3 < 0 Thus 𝑑2𝑣𝑑 𝑥2<0 when 𝑥= 𝑅3 ⇒ Volume is Maximum when 𝑥= 𝑅3 Finding maximum volume From (1) Volume of cone = 13𝜋 𝑅2− 𝑥2 𝑅+𝑥 Putting 𝑥 = 𝑅3 = 13𝜋 𝑅2− 𝑅32 𝑅+ 𝑅3 = 13𝜋 𝑅2− 𝑅29 3𝑅 + 𝑅3 = 13𝜋 9 𝑅2− 𝑅29 4𝑅3 = 𝟑𝟐𝟖𝟏𝝅 𝑹𝟑 Hence proved

Ex 6.5

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Ex 6.5,23 Important You are here

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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.