# Misc 18

Last updated at Dec. 27, 2017 by Teachoo

Last updated at Dec. 27, 2017 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 α is one-third that of the cone and the greatest volume of cylinder is 427 𝜋ℎ3 tan2 𝛼 Let PQR be the cone of height h i.e. PO = h & α be the semi vertical angle of cone Let 𝑥 be the radius of cylinder ABCD which is inscribed in the cone PQR Now, Height of cylinder = OO’ = PO – PO’ In ∆AP𝑂 tan α = 𝐴𝑂′𝑃𝑂′ tan α = 𝑥𝑉𝑂′ VO’ = 𝑥tanα VO’ = 𝑥 cotα Now Height of cylinder = OO’ OO’ = VO – VO’ OO’ = h – 𝑥 cot α We need to maximize volume of cylinder Let V be the volume of cylinder V = π 𝑟𝑎𝑑𝑖𝑢𝑠 2ℎ𝑒𝑖𝑔ℎ𝑡 V = π 𝐴′𝑂′2𝑂 𝑂′ V = π 𝑥2ℎ−𝑥cotα V = π 𝑥2 ℎ−𝑥α V = π h𝑥2 – π cot α. 𝑥3 Differentiating w.r.t 𝑥 𝑑𝑉𝑑𝑥=𝑑𝜋ℎ𝑥2−𝜋cotα.𝑥3𝑑𝑥 𝑑𝑉𝑑𝑥= π h𝑑𝑥2𝑑𝑥−𝜋cotα.𝑑𝑥3𝑑𝑥 𝑑𝑉𝑑𝑥= πh. 2𝑥 – π cot α. 3𝑥2 𝑑𝑉𝑑𝑥= 2πh𝑥 – 3π cot α 𝑥2 Putting 𝑑𝑉𝑑𝑥= 0 2π h 𝑥 – 3π cot α 𝑥2 = 0 3π cot α 𝑥2 = 2π h 𝑥 𝑥 = 2𝜋ℎ 𝑥3𝜋cot α.𝑥 𝑥 = 2ℎ3cot α 𝑥 = 2ℎ3tanα Now finding 𝑑2𝑉𝑑𝑥2 𝑑2𝑉𝑑𝑥2= 𝑑2𝜋 ℎ𝑥 − 3𝜋 𝑐𝑜𝑡α . 𝑥2𝑑𝑥 𝑑2𝑉𝑑𝑥= 2π h – 3π cot α . 2𝑥 𝑑2𝑣𝑑𝑥2=2π h – 6π cot α . 𝑥 Putting value of 𝑥 = 2ℎ3𝑐𝑜𝑡α 𝑑2𝑣𝑑𝑥2=2𝜋ℎ−4𝜋ℎ = –2π h < 0 ⇒ 𝑑2𝑣𝑑𝑥2 < 0 at 𝑥 = 2ℎ3cot α ⇒ Hence, V is maximum when 𝑥 = 2ℎ3cotα From (1) OO’ = h – 𝑥 cot α OO’ = h – 2ℎ3cotα ×α OO’ = h – 2ℎ3 OO’ = ℎ 3 Thus, Volume is maximum when Height of cylinder is = 13 ×ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑛𝑒 Finding Maximum Volume V = π 𝑥2 ℎ −𝑥cotα V = π 2ℎ3cotα2ℎ−2ℎ3cotα ×cotα V = π 4ℎ29 cot2αℎ−2ℎ3 V = π 4ℎ29cot2αℎ3 V = 4 9 ×3𝜋ℎ2 ×ℎcot2α V = 427𝜋ℎ3𝜋ℎ3cot2α V = 42𝜋𝜋ℎ3.tan2α Hence height of cylinder is one third of cone & greatest volume of cylinder is 𝟒𝟐𝝅𝝅𝒉𝟑𝐭𝐚𝐧𝟐𝜶

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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