# Misc 18 - 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 18 Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle Ξ± is one-third that of the cone and the greatest volume of cylinder is 4/27 πβ3 tan2 πΌ Let PQR be the cone of height h i.e. PO = h & Ξ± be the semi vertical angle of cone Let π₯ be the radius of cylinder ABCD which is inscribed in the cone PQR Now, Height of cylinder = OOβ = PO β POβ In βAPπβ tan Ξ± = (π΄π^β²)/(ππ^β² ) tan Ξ± = π₯/(ππ^β² ) POβ = π₯/tanβ‘Ξ±" " POβ = π₯ cotΞ± Now Height of cylinder = OOβ OOβ = PO β POβ (Since AOβ = DO = Radius of cylinder) OOβ = h β π₯ cot Ξ± We need to maximize volume of cylinder Let V be the volume of cylinder V = Ο (πππππ’π )^2 (βπππβπ‘) V = Ο (π΄^β² π^β² )^2 (π πβ²) V = Ο (π₯)^2 (ββπ₯ cotβ‘Ξ± ) V = Ο π₯2 (ββπ₯Ξ±) V = Ο hπ₯2 β Ο cot Ξ±. π₯3 Differentiating w.r.t π₯ ππ/ππ₯=π(πβπ₯^2βπ cotβ‘γΞ±.π₯^3 γ )/ππ₯ (From (1)) ππ/ππ₯= Ο h(π(π₯)^2)/ππ₯βπ cotβ‘γΞ±.(π(π₯)^3)/ππ₯γ ππ/ππ₯= Οh. 2π₯ β Ο cot Ξ±. 3π₯2 ππ/ππ₯= 2Οhπ₯ β 3Ο cot Ξ± π₯2 Putting ππ/ππ₯= 0 2Ο h π₯ β 3Ο cot Ξ± π₯2 = 0 3Ο cot Ξ± π₯2 = 2Ο h π₯ π₯ = (2πβ π₯)/(3π cotβ‘γ Ξ±.π₯γ ) π₯ = 2β/(3 cotβ‘γ Ξ±γ ) π₯ = 2β/3 tanβ‘Ξ± Now finding (π^2 π)/(ππ₯^2 ) (π^2 π)/(ππ₯^2 )= π(2π βπ₯ β 3π πππ‘Ξ± . γ π₯γ^2 )/ππ₯ (π^2 π)/ππ₯= 2Ο h β 3Ο cot Ξ± . 2π₯ (π^2 π£)/(ππ₯^2 )=2Ο h β 6Ο cot Ξ± . π₯ Putting value of π₯ = 2β/(3 πππ‘β‘Ξ± ) (π^2 π£)/(ππ₯^2 )=2πββ4πβ = β2Ο h < 0 Thus, (π^2 π£)/(ππ₯^2 ) < 0 at π₯ = 2β/(3 cotβ‘γ Ξ±γ ) Hence, V is maximum when π₯ = 2β/(3 cotβ‘Ξ± ) From (1) OOβ = h β π₯ cot Ξ± OOβ = h β 2β/(3 cotβ‘Ξ±" " ) Γcotβ‘Ξ± OOβ = h β 2β/3 OOβ = β/( 3) Thus, Volume is maximum when Height of cylinder is = 1/3 Γ π»πππβπ‘ ππ ππππ Finding Maximum Volume V = Ο π₯2 (β βπ₯ cotβ‘Ξ± ) V = Ο (2β/(3 cotβ‘Ξ± ))^2 (ββ2β/(3 cotβ‘Ξ± ) Γcotβ‘Ξ± ) V = Ο ((4β^2)/(9 γ cotγ^2β‘Ξ± ))(ββ2β/3) V = Ο ((4β^2)/(9 cot^2β‘Ξ± ))(β/3) V = (4 )/(9 Γ3) ((πβ^2 Γβ)/cot^2β‘Ξ± ) V = 4/27 ((πβ^3)/cot^2β‘Ξ± ) V = 4/27 πβ^3.tan^2β‘Ξ± Hence height of cylinder is one third of cone & greatest volume of cylinder is π/ππ π π^π γπππ§γ^πβ‘πΆ

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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.