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Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at April 19, 2021 by Teachoo

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.Given Radius of sphere = R Let h be the height & π be the diameter of cylinder In β π¨π©πͺ 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 π2 = 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)^2Γ β V = Ο Γ π₯^2/4Γ β V = Ο ((4π ^2 β β^2 ))/4 Γ β V = (4π ^2 πβ)/4β(πβ^3)/4 V = ΟhR2 β (π π^π)/π Differentiating w.r.t π ππ/πβ=π(πβπ ^2 β πβ^3/4)/πβ ππ/πβ= ΟR2 π(β)/πββπ/4 π(β^3 )/πβ ππ΅/πβ= ΟR2 β π/4 (3β^2 ) ππ/πβ= ΟR2 β 3π/4 h2 Putting π π½/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 = ππΉ/βπ Finding (π ^π π½)/(π π^π ) ππ/πβ=ππ ^2β3/(4 ) π β^2 Differentiating w.r.t. h (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 ^π π½)/(π π^π )=(βππ π)/π Since (π ^π π½)/(π π^π )<π for h = 2π /β3 β΄ Volume is maximum for h = 2π /β3 We also need to find Maximum Volume V = ΟhR2 β (πβ^3)/4 V = ΟR2 Γ 2π /β3 β π/4 Γ (2π /β3)^3 V = (2ππ ^3)/β3 β π/4 Γ(8π ^3)/(3β3) V = (2ππ ^3)/β3 β (2ππ ^3)/(3β3) V = (2ππ ^3)/β3 (1β1/3) V = (2ππ ^3)/β3 Γ2/3 V = (ππ πΉ^π)/(πβπ) cubic unit