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Misc 14 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2𝑅/√3 . Also find the maximum volume.Given Radius of sphere = R Let h be the height & 𝒙 be the diameter of cylinder In βˆ† 𝑨𝑩π‘ͺ Using Pythagoras theorem (𝐢𝐡)^2+(𝐴𝐡)^2=(𝐴𝐢)^2 h2 + π‘₯^2=(𝑅+𝑅)^2 h2 + π‘₯2 =(2𝑅)^2 h2 + π‘₯2 = 4R2 𝒙2 = 4R2 – h2 We need to find maximum volume of cylinder Let V be the volume of cylinder V = Ο€ (π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ )^2Γ—(β„Žπ‘’π‘–π‘”β„Žπ‘‘) V = Ο€ (π‘₯/2)^2Γ— β„Ž V = Ο€ Γ— π‘₯^2/4Γ— β„Ž V = Ο€ ((4𝑅^2 βˆ’ β„Ž^2 ))/4 Γ— β„Ž V = (4𝑅^2 πœ‹β„Ž)/4βˆ’(πœ‹β„Ž^3)/4 V = Ο€hR2 – (𝝅𝒉^πŸ‘)/πŸ’ Differentiating w.r.t 𝒉 𝑑𝑉/π‘‘β„Ž=𝑑(πœ‹β„Žπ‘…^2 βˆ’ πœ‹β„Ž^3/4)/π‘‘β„Ž 𝑑𝑉/π‘‘β„Ž= Ο€R2 𝑑(β„Ž)/π‘‘β„Žβˆ’πœ‹/4 𝑑(β„Ž^3 )/π‘‘β„Ž 𝑑𝐡/π‘‘β„Ž= Ο€R2 – πœ‹/4 (3β„Ž^2 ) 𝑑𝑉/π‘‘β„Ž= Ο€R2 – 3πœ‹/4 h2 Putting 𝒅𝑽/𝒅𝒉=𝟎 Ο€ R2 – 3/4 πœ‹ β„Ž^2=0 3/4 πœ‹β„Ž^2=πœ‹π‘…^2 h2 = (πœ‹π‘…^2)/(3/4 πœ‹) h2 = (4𝑅^2)/3 h =√((4𝑅^2)/3) h = πŸπ‘Ή/βˆšπŸ‘ Finding (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 ) 𝑑𝑉/π‘‘β„Ž=πœ‹π‘…^2βˆ’3/(4 ) πœ‹ β„Ž^2 Differentiating w.r.t. h (𝑑^2 𝑉)/(π‘‘β„Ž^2 )= 𝑑(πœ‹π‘…^2 βˆ’ 3/4 πœ‹β„Ž^2 )/π‘‘β„Ž (𝑑^2 𝑉)/(π‘‘β„Ž^2 )= 0 – 3πœ‹/4 Γ—2β„Ž (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 )=(βˆ’πŸ‘π…π’‰)/𝟐 Since (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 )<𝟎 for h = 2𝑅/√3 ∴ Volume is maximum for h = 2𝑅/√3 We also need to find Maximum Volume V = Ο€hR2 – (πœ‹β„Ž^3)/4 V = Ο€R2 Γ— 2𝑅/√3 – πœ‹/4 Γ— (2𝑅/√3)^3 V = (2πœ‹π‘…^3)/√3 – πœ‹/4 Γ—(8𝑅^3)/(3√3) V = (2πœ‹π‘…^3)/√3 – (2πœ‹π‘…^3)/(3√3) V = (2πœ‹π‘…^3)/√3 (1βˆ’1/3) V = (2πœ‹π‘…^3)/√3 Γ—2/3 V = (πŸ’π…π‘Ή^πŸ‘)/(πŸ‘βˆšπŸ‘) cubic unit

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.