Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


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 πœ‹β„Ž3 tan2 𝛼 Let PQR be the cone of height h i.e. PO = h & Ξ± be the semi vertical angle of cone Let π‘₯ be the radius of cylinder ABCD which is inscribed in the cone PQR Now, Height of cylinder = OO’ = PO – PO’ In βˆ†AP𝑂’ tan Ξ± = (𝐴𝑂^β€²)/(𝑃𝑂^β€² ) tan Ξ± = π‘₯/(𝑃𝑂^β€² ) PO’ = π‘₯/tan⁑α" " PO’ = π‘₯ cotΞ± Now Height of cylinder = OO’ OO’ = PO – PO’ (Since AO’ = DO = Radius of cylinder) OO’ = h – π‘₯ cot Ξ± We need to maximize volume of cylinder Let V be the volume of cylinder V = Ο€ (π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  )^2 (β„Žπ‘’π‘–π‘”β„Žπ‘‘) V = Ο€ (𝐴^β€² 𝑂^β€² )^2 (𝑂 𝑂′) V = Ο€ (π‘₯)^2 (β„Žβˆ’π‘₯ cot⁑α ) V = Ο€ π‘₯2 (β„Žβˆ’π‘₯Ξ±) V = Ο€ hπ‘₯2 – Ο€ cot Ξ±. π‘₯3 Differentiating w.r.t π‘₯ 𝑑𝑉/𝑑π‘₯=𝑑(πœ‹β„Žπ‘₯^2βˆ’πœ‹ cot⁑〖α.π‘₯^3 γ€— )/𝑑π‘₯ (From (1)) 𝑑𝑉/𝑑π‘₯= Ο€ h(𝑑(π‘₯)^2)/𝑑π‘₯βˆ’πœ‹ cot⁑〖α.(𝑑(π‘₯)^3)/𝑑π‘₯γ€— 𝑑𝑉/𝑑π‘₯= Ο€h. 2π‘₯ – Ο€ cot Ξ±. 3π‘₯2 𝑑𝑉/𝑑π‘₯= 2Ο€hπ‘₯ – 3Ο€ cot Ξ± π‘₯2 Putting 𝑑𝑉/𝑑π‘₯= 0 2Ο€ h π‘₯ – 3Ο€ cot Ξ± π‘₯2 = 0 3Ο€ cot Ξ± π‘₯2 = 2Ο€ h π‘₯ π‘₯ = (2πœ‹β„Ž π‘₯)/(3πœ‹ cot⁑〖 Ξ±.π‘₯γ€— ) π‘₯ = 2β„Ž/(3 cot⁑〖 Ξ±γ€— ) π‘₯ = 2β„Ž/3 tan⁑α Now finding (𝑑^2 𝑉)/(𝑑π‘₯^2 ) (𝑑^2 𝑉)/(𝑑π‘₯^2 )= 𝑑(2πœ‹ β„Žπ‘₯ βˆ’ 3πœ‹ π‘π‘œπ‘‘Ξ± . γ€– π‘₯γ€—^2 )/𝑑π‘₯ (𝑑^2 𝑉)/𝑑π‘₯= 2Ο€ h – 3Ο€ cot Ξ± . 2π‘₯ (𝑑^2 𝑣)/(𝑑π‘₯^2 )=2Ο€ h – 6Ο€ cot Ξ± . π‘₯ Putting value of π‘₯ = 2β„Ž/(3 π‘π‘œπ‘‘β‘Ξ± ) (𝑑^2 𝑣)/(𝑑π‘₯^2 )=2πœ‹β„Žβˆ’4πœ‹β„Ž = –2Ο€ h < 0 Thus, (𝑑^2 𝑣)/(𝑑π‘₯^2 ) < 0 at π‘₯ = 2β„Ž/(3 cot⁑〖 Ξ±γ€— ) Hence, V is maximum when π‘₯ = 2β„Ž/(3 cot⁑α ) From (1) OO’ = h – π‘₯ cot Ξ± OO’ = h – 2β„Ž/(3 cot⁑α" " ) Γ—cot⁑α OO’ = h – 2β„Ž/3 OO’ = β„Ž/( 3) Thus, Volume is maximum when Height of cylinder is = 1/3 Γ— π»π‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘π‘œπ‘›π‘’ Finding Maximum Volume V = Ο€ π‘₯2 (β„Ž βˆ’π‘₯ cot⁑α ) V = Ο€ (2β„Ž/(3 cot⁑α ))^2 (β„Žβˆ’2β„Ž/(3 cot⁑α ) Γ—cot⁑α ) V = Ο€ ((4β„Ž^2)/(9 γ€– cotγ€—^2⁑α ))(β„Žβˆ’2β„Ž/3) V = Ο€ ((4β„Ž^2)/(9 cot^2⁑α ))(β„Ž/3) V = (4 )/(9 Γ—3) ((πœ‹β„Ž^2 Γ—β„Ž)/cot^2⁑α ) V = 4/27 ((πœ‹β„Ž^3)/cot^2⁑α ) V = 4/27 πœ‹β„Ž^3.tan^2⁑α Hence height of cylinder is one third of cone & greatest volume of cylinder is πŸ’/πŸπŸ• 𝝅𝒉^πŸ‘ γ€–π­πšπ§γ€—^𝟐⁑𝜢

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.