Misc 18 - Chapter 6 Class 12 Application of Derivatives

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