# Misc 18 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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 is one-third that of the cone and the greatest volume of cylinder is 4 27 3 tan2 Let PQR be the cone of height h i.e. PO = h & be the semi vertical angle of cone Let be the radius of cylinder ABCD which is inscribed in the cone PQR Now, Height of cylinder = OO = PO PO In AP tan = tan = VO = tan VO = cot Now Height of cylinder = OO OO = VO VO OO = h cot We need to maximize volume of cylinder Let V be the volume of cylinder V = 2 V = 2 V = 2 cot V = 2 V = h 2 cot . 3 Differentiating w.r.t = 2 cot . 3 = h 2 cot . 3 = h. 2 cot . 3 2 = 2 h 3 cot 2 Putting = 0 2 h 3 cot 2 = 0 3 cot 2 = 2 h = 2 3 cot . = 2 3 cot = 2 3 tan Now finding 2 2 2 2 = 2 3 . 2 2 = 2 h 3 cot . 2 2 2 =2 h 6 cot . Putting value of = 2 3 2 2 =2 4 = 2 h < 0 2 2 < 0 at = 2 3 cot Hence, V is maximum when = 2 3 cot From (1) OO = h cot OO = h 2 3 cot OO = h 2 3 OO = 3 Thus, Volume is maximum when Height of cylinder is = 1 3 Finding Maximum Volume V = 2 cot V = 2 3 cot 2 2 3 cot cot V = 4 2 9 cot 2 2 3 V = 4 2 9 cot 2 3 V = 4 9 3 2 cot 2 V = 4 27 3 3 cot 2 V = 4 2 3 . tan 2 Hence height of cylinder is one third of cone & greatest volume of cylinder is

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.