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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 Ξ” BOC is a right angled triangle Using Pythagoras theorem in βˆ†BOC 〖𝑂𝐡〗^2=〖𝐡𝐢〗^2+〖𝑂𝐢〗^2 R2 =〖𝐡𝐢〗^2+π‘₯^2 BC2 = 𝑅^2 – x2 BC = √(𝑅^2βˆ’π‘₯^2 ) Thus, Radius of cone = BC = √(𝑅^2βˆ’π‘₯^2 ) Height of cone = OC + OA = R + x We need to show Maximum volume of cone = 8/27 Γ— Volume of sphere = 8/27 Γ— 4/3 πœ‹π‘…^3 = 32/81 πœ‹π‘…^3 Let V be the volume of a cone We know that Volume of a cone = 1/3 πœ‹(π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  )^2 (β„Žπ‘’π‘–π‘”β„Žπ‘‘ ) V = 1/3 πœ‹(√(𝑅^2βˆ’π‘₯^2 ))^2 (𝑅+π‘₯) V = 1/3 πœ‹(𝑅^2βˆ’π‘₯^2 )(𝑅+π‘₯) V = 1/3 πœ‹(𝑅^2 (𝑅+π‘₯)βˆ’π‘₯^2 (𝑅+π‘₯)) V = 1/3 πœ‹(𝑅^(3 )+𝑅^2 π‘₯βˆ’π‘₯^2 π‘…βˆ’π‘₯^3 ) Diff w.r.t 𝒙 𝑑𝑉/𝑑π‘₯=𝑑/𝑑π‘₯ [1/3 πœ‹(𝑅^(3 )+𝑅^2 π‘₯βˆ’π‘₯^2 π‘…βˆ’π‘₯^3 )] 𝑑𝑉/𝑑π‘₯=πœ‹/3 [𝑑/𝑑π‘₯ (𝑅^(3 )+𝑅^2 π‘₯βˆ’π‘₯^2 π‘…βˆ’π‘₯^3 )] 𝑑𝑉/𝑑π‘₯=πœ‹/3 (0+𝑅^2.1βˆ’π‘… Γ—2π‘₯βˆ’3π‘₯^2 ) 𝑑𝑉/𝑑π‘₯=πœ‹/3 (𝑅^2βˆ’2𝑅π‘₯βˆ’3π‘₯^2 ) Putting 𝒅𝑽/𝒅𝒙=𝟎 1/3 πœ‹(𝑅^2βˆ’2𝑅π‘₯βˆ’3π‘₯^2 )=0 𝑅^2βˆ’2𝑅π‘₯βˆ’3π‘₯^2=0 𝑅^2βˆ’3𝑅π‘₯+𝑅π‘₯βˆ’3π‘₯^2=0 R(π‘…βˆ’3π‘₯)+π‘₯(π‘…βˆ’3π‘₯)=0 (𝑅+π‘₯)(π‘…βˆ’3π‘₯)=0 So, x = –R & π‘₯=𝑅/3 Since x cannot be negative π‘₯=𝑅/3 Finding (𝒅^𝟐 𝒗)/(𝒅𝒙^𝟐 ) 𝑑𝑣/𝑑π‘₯=1/3 πœ‹(𝑅^2βˆ’2𝑅π‘₯βˆ’3π‘₯^2 ) (𝑑^2 𝑣)/(𝑑π‘₯^2 )=πœ‹/3 𝑑/𝑑π‘₯ (𝑅^2βˆ’2𝑅π‘₯βˆ’3π‘₯^2 ) (𝑑^2 𝑣)/(𝑑π‘₯^2 )=πœ‹/3 (0βˆ’2π‘…βˆ’6π‘₯) (𝑑^2 𝑣)/(𝑑π‘₯^2 )=(βˆ’πœ‹)/3 (2𝑅+6π‘₯) Putting 𝒙=𝑹/πŸ‘ γ€–(𝑑^2 𝑣)/(𝑑π‘₯^2 )β”‚γ€—_(π‘₯ = 𝑅/3) =(βˆ’πœ‹)/3 (2𝑅+6(𝑅/3)) =(βˆ’πœ‹)/3 (2𝑅+2𝑅) =(βˆ’πœ‹)/3 (4𝑅) =(βˆ’4πœ‹π‘…)/3 < 0 Thus (𝑑^2 𝑣)/(𝑑π‘₯^2 )<0 when π‘₯=𝑅/3 ∴ Volume is Maximum when π‘₯=𝑅/3 Finding maximum volume From (1) Volume of cone = 1/3 πœ‹(𝑅^2βˆ’π‘₯^2 )(𝑅+π‘₯) Putting π‘₯ = 𝑅/3 = 1/3 πœ‹(𝑅^2βˆ’(𝑅/3)^2 )(𝑅+𝑅/3) = 1/3 πœ‹(𝑅^2βˆ’π‘…^2/9)((3𝑅 + 𝑅)/3) = 1/3 πœ‹((9𝑅^2βˆ’ 𝑅^2)/9)(4𝑅/3) = πŸ‘πŸ/πŸ–πŸ 𝝅𝑹^πŸ‘ Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.