Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Misc 17 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 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.