Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Misc 14 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2π/β3 . 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©πͺ Using Pythagoras theorem (πΆπ΅)^2+(π΄π΅)^2=(π΄πΆ)^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 (πππππ’π )^2Γ(βπππβπ‘) 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 πβ)/4β(πβ^3)/4 V = ΟhR2 β (ππ^π)/π Differentiating w.r.t π ππ/πβ=π(πβπ^2 β πβ^3/4)/πβ ππ/πβ= ΟR2 π(β)/πββπ/4 π(β^3 )/πβ ππ΅/πβ= ΟR2 β π/4 (3β^2 ) ππ/πβ= ΟR2 β 3π/4 h2 Putting ππ½/ππ=π Ο R2 β 3/4 π β^2=0 3/4 πβ^2=ππ^2 h2 = (ππ^2)/(3/4 π) h2 = (4π^2)/3 h =β((4π^2)/3) h = ππΉ/βπ Finding (π^π π½)/(ππ^π ) ππ/πβ=ππ^2β3/(4 ) π β^2 Differentiating w.r.t. h (π^2 π)/(πβ^2 )= π(ππ^2 β 3/4 πβ^2 )/πβ (π^2 π)/(πβ^2 )= 0 β 3π/4 Γ2β (π^π π½)/(ππ^π )=(βπππ)/π Since (π^π π½)/(ππ^π )<π for h = 2π/β3 β΄ Volume is maximum for h = 2π/β3 We also need to find Maximum Volume V = ΟhR2 β (πβ^3)/4 V = ΟR2 Γ 2π/β3 β π/4 Γ (2π/β3)^3 V = (2ππ^3)/β3 β π/4 Γ(8π^3)/(3β3) V = (2ππ^3)/β3 β (2ππ^3)/(3β3) V = (2ππ^3)/β3 (1β1/3) V = (2ππ^3)/β3 Γ2/3 V = (πππΉ^π)/(πβπ) cubic unit