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Chapter 6 Class 12 Application of Derivatives (Term 1)

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Last updated at April 19, 2021 by Teachoo

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 πΌGiven Height of cone = h Semi-vertical angle of cone = πΆ Let Radius of Cylinder = π 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 cotπΆ Now Height of cylinder = OOβ = PO β POβ = h β π cot πΆ 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 = π ππ^πβπ πππβ‘πΆ π^π Differentiating w.r.t π₯ π π½/π π=π(πβπ₯^2 β π cotβ‘Ξ± π₯^3 )/ππ₯ 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 π π½/π π= 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 = ππ/(π πππβ‘γ πΆγ ) Now finding (π ^π π½)/(π π^π ) (π^2 π)/(ππ₯^2 )= π(2π βπ₯ β 3π πππ‘Ξ± . γ π₯γ^2 )/ππ₯ (π^2 π)/ππ₯= 2Οh β 3Ο cot Ξ± . 2π₯ (π^2 π)/(ππ₯^2 )= 2Οh β 6Ο cot Ξ± . π₯ Putting value of π₯ = 2β/(3 πππ‘β‘Ξ± ) (π^2 π)/(ππ₯^2 )= 2Οh β 6Ο cot Ξ± Γ 2β/(3 πππ‘β‘Ξ± ) (π^2 π)/(ππ₯^2 )= 2Οh β 4Οh (π^2 π)/(ππ₯^2 )= β2Οh Since (π ^π π½)/(π π^π )<π for π₯ = 2β/(3 πππ‘β‘Ξ± ) β΄ Volume is maximum for π₯ = 2β/(3 πππ‘β‘Ξ± ) We need to find Height and Volume For Height Height of cylinder = π β π cot πΆ = β β cot πΌ Γ 2β/(3 πππ‘β‘γ πΌγ ) = β β 2β/3 = π/π Hence, Height of cylinder is one third of cone 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 = 4/27 ((πβ^3)/cot^2β‘Ξ± ) V = π/ππ π π^π.γπππγ^πβ‘πΆ Thus, greatest volume of cylinder is 4/27 πβ^3 tan^2β‘πΌ