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

Transcript

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. Let R be the radius of sphere Let h be the height & π‘₯ be the diameter of cylinder In βˆ† 𝐴𝐡𝐢 Using Pythagoras theorem (𝐢𝐡)^2+(𝐴𝐡)^2=(𝐴𝐢)^2 h2 + (π‘₯)^2=(𝑅+𝑅)^2 h2 + π‘₯2 =(2𝑅)^2 h2 + π‘₯2 = 4R2 π‘₯2 = 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 …(1) 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 – (πœ‹β„Ž^3)/4 Different w.r.t β„Ž 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 𝑑𝑣/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 = 2𝑅/√3 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.x (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)/(π‘‘β„Ž^2 )= (βˆ’3πœ‹β„Ž)/2 Putting h = 2𝑅/√3 (𝑑^2 𝑣)/(π‘‘β„Ž^2 ) = (βˆ’3πœ‹)/2 (2𝑅/√3) = β€“βˆš3πœ‹π‘… < 0 ∴ h = 2𝑅/√3 is point of maxima So, volume is maximum when h = 2𝑅/√3 From (1) π‘₯2 = 4R2 – h2 π‘₯2 = 4R2 – (2𝑅/√3)^2 π‘₯2 = 4R2 – (4𝑅^2)/3 π‘₯2 = (12𝑅^2 βˆ’ 4𝑅^2)/3 π‘₯2 = 8/(3 ) γ€– 𝑅〗^2 Maximum value of volume is V = Ο€ (π‘₯/2)^2 β„Ž V = πœ‹/4 π‘₯^2 β„Ž Putting value of π‘₯2 & h V = πœ‹/4 Γ— (8𝑅^2)/3 Γ— 2𝑅/√3 V = ( 16πœ‹π‘…^3)/(12√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 9 years. He provides courses for Maths and Science at Teachoo.