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Last updated at Jan. 7, 2020 by Teachoo

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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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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.