





Miscellaneous
Misc 1 (b) Important Deleted for CBSE Board 2022 Exams
Misc 2 Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6 Important
Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13 Important
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important You are here
Misc. 19 (MCQ) Deleted for CBSE Board 2022 Exams
Misc 20 (MCQ) Important
Misc 21 (MCQ) Important
Misc 22 (MCQ)
Misc. 23 (MCQ) Important
Misc 24 (MCQ) Important
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