# Misc 17 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Misc 17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2π /β3 . Also find the maximum volume. Let R be the radius of sphere Let h be the height & π₯ be the diameter of cylinder In β π΄π΅πΆ Using Pythagoras theorem (πΆπ΅)^2+(π΄π΅)^2=(π΄πΆ)^2 h2 + (π₯)^2=(π +π )^2 h2 + π₯2 =(2π )^2 h2 + π₯2 = 4R2 π₯2 = 4R2 β h2 We need to find maximum volume of cylinder Let V be the volume of cylinder V = Ο (πππππ’π )^2Γ(βπππβπ‘) V = Ο (π₯/2)^2Γβ V = Ο Γ π₯^2/4Γβ V = Ο ((4π ^2 β β^2 ))/4 Γβ β¦(1) V = Ο ((4π ^2 β β^2 ))/4 Γβ V = (4π ^2 πβ)/4β(πβ^3)/4 V = ΟhR2 β (πβ^3)/4 Different w.r.t β ππ£/πβ=π(πβπ ^2 β πβ^3/4)/πβ ππ£/πβ= Ο R2 π(β)/πββπ/4 π(β^3 )/πβ ππ£/πβ= Ο R2 β π/4 (3β^2 ) ππ£/πβ= Ο R2 β 3π/4 h2 Putting π π/π π=π Ο R2 β 3/4 π β^2=0 3/4 πβ^2=ππ ^2 h2 = (ππ ^2)/(3/4 π) h2 = (4π ^2)/3 h =β((4π ^2)/3) h = 2π /β3 Finding (π ^π π)/(π π^π ) ππ£/πβ=ππ ^2β3/(4 ) π β^2 Differentiating w.r.t.x (π^2 π£)/(πβ^2 )= π(ππ ^2 β 3/4 πβ^2 )/πβ (π^2 π£)/(πβ^2 )= 0 β 3π/4 Γ2β (π^2 π£)/(πβ^2 )= (β3πβ)/2 Putting h = 2π /β3 (π^2 π£)/(πβ^2 ) = (β3π)/2 (2π /β3) = ββ3ππ < 0 β΄ h = 2π /β3 is point of maxima So, volume is maximum when h = 2π /β3 From (1) π₯2 = 4R2 β h2 π₯2 = 4R2 β (2π /β3)^2 π₯2 = 4R2 β (4π ^2)/3 π₯2 = (12π ^2 β 4π ^2)/3 π₯2 = 8/(3 ) γ π γ^2 Maximum value of volume is V = Ο (π₯/2)^2 β V = π/4 π₯^2 β Putting value of π₯2 & h V = π/4 Γ (8π ^2)/3 Γ 2π /β3 V = ( 16ππ ^3)/(12β3) V = (ππ πΉ^π)/(πβπ) cubic unit

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Chapter 6 Class 12 Application of Derivatives

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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.