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Last updated at Jan. 7, 2020 by Teachoo

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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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Misc 17 Important You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.