Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Misc 15 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 πβ3 tan2 πΌGiven Height of cone = h Semi-vertical angle of cone = πΆ Let Radius of Cylinder = π Now, Height of cylinder = OOβ = PO β POβ In βAPπβ tan Ξ± = (π΄π^β²)/(ππ^β² ) tan Ξ± = π₯/(ππ^β² ) POβ = π₯/tanβ‘Ξ±" " POβ = π cotπΆ Now Height of cylinder = OOβ = PO β POβ = h β π cot πΆ We need to maximize volume of cylinder Let V be the volume of cylinder V = Ο (πππππ’π  )^2 (βπππβπ‘) V = Ο (π΄^β² π^β² )^2 (π πβ²) V = Ο π₯^2 (ββπ₯ cotβ‘Ξ± ) V = πππ^πβπ πππβ‘πΆ π^π Differentiating w.r.t π₯ ππ½/ππ=π(πβπ₯^2 β π cotβ‘Ξ± π₯^3 )/ππ₯ ππ/ππ₯= Ο h(π(π₯)^2)/ππ₯βπ cotβ‘γΞ±.(π(π₯)^3)/ππ₯γ ππ/ππ₯= Οh. 2π₯ β Ο cot Ξ±. 3π₯2 ππ/ππ₯= 2Οhπ₯ β 3Ο cot Ξ± π₯2 Putting ππ½/ππ= 0 2Ο h π₯ β 3Ο cot Ξ± π₯2 = 0 3Ο cot Ξ± π₯2 = 2Ο h π₯ π₯ = (2πβ π₯)/(3π cotβ‘γ Ξ±.π₯γ ) π = ππ/(π πππβ‘γ πΆγ ) Now finding (π^π π½)/(ππ^π ) (π^2 π)/(ππ₯^2 )= π(2π βπ₯ β 3π πππ‘Ξ± . γ π₯γ^2 )/ππ₯ (π^2 π)/ππ₯= 2Οh β 3Ο cot Ξ± . 2π₯ (π^2 π)/(ππ₯^2 )= 2Οh β 6Ο cot Ξ± . π₯ Putting value of π₯ = 2β/(3 πππ‘β‘Ξ± ) (π^2 π)/(ππ₯^2 )= 2Οh β 6Ο cot Ξ± Γ 2β/(3 πππ‘β‘Ξ± ) (π^2 π)/(ππ₯^2 )= 2Οh β 4Οh (π^2 π)/(ππ₯^2 )= β2Οh Since (π^π π½)/(ππ^π )<π for π₯ = 2β/(3 πππ‘β‘Ξ± ) β΄ Volume is maximum for π₯ = 2β/(3 πππ‘β‘Ξ± ) We need to find Height and Volume For Height Height of cylinder = π β π cot πΆ = β β cot πΌ Γ 2β/(3 πππ‘β‘γ πΌγ ) = β β 2β/3 = π/π Hence, Height of cylinder is one third of cone Finding Maximum Volume V = Ο π₯2 (β βπ₯ cotβ‘Ξ± ) V = Ο (2β/(3 cotβ‘Ξ± ))^2 (ββ2β/(3 cotβ‘Ξ± ) Γcotβ‘Ξ± ) V = Ο ((4β^2)/(9 γ cotγ^2β‘Ξ± ))(ββ2β/3) V = Ο ((4β^2)/(9 cot^2β‘Ξ± ))(β/3) V = 4/27 ((πβ^3)/cot^2β‘Ξ± ) V = π/ππ ππ^π.γπππγ^πβ‘πΆ Thus, greatest volume of cylinder is 4/27 πβ^3 tan^2β‘πΌ

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.