Ex 6.3,23 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
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Ex 6.3,7 Important
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Ex 6.3,9 Important
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Ex 6.3,18 Important
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Ex 6.3,22 Important
Ex 6.3,23 Important You are here
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 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