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Misc 18 - Show that height of cylinder of greatest volume

Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 4 Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 5 Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 6 Misc 18 - Chapter 6 Class 12 Application of Derivatives - Part 7


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 Ξ± is one-third that of the cone and the greatest volume of cylinder is 4/27 πœ‹β„Ž3 tan2 𝛼Given Height of cone = h Semi-vertical angle of cone = 𝜢 Let Radius of Cylinder = 𝒙 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’ = 𝒙 cot𝜢 Now Height of cylinder = OO’ = PO – PO’ = h – 𝒙 cot 𝜢 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 = 𝝅𝒉𝒙^πŸβˆ’π… π’„π’π’•β‘πœΆ 𝒙^πŸ‘ Differentiating w.r.t π‘₯ 𝒅𝑽/𝒅𝒙=𝑑(πœ‹β„Žπ‘₯^2 βˆ’ πœ‹ cot⁑α π‘₯^3 )/𝑑π‘₯ 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/𝒅𝒙= 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€— ) 𝒙 = πŸπ’‰/(πŸ‘ 𝒄𝒐𝒕⁑〖 πœΆγ€— ) Now finding (𝒅^𝟐 𝑽)/(𝒅𝒙^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 )= 2Ο€h – 6Ο€ cot Ξ± Γ— 2β„Ž/(3 π‘π‘œπ‘‘β‘Ξ± ) (𝑑^2 𝑉)/(𝑑π‘₯^2 )= 2Ο€h – 4Ο€h (𝑑^2 𝑉)/(𝑑π‘₯^2 )= –2Ο€h Since (𝒅^𝟐 𝑽)/(𝒅𝒙^𝟐 )<𝟎 for π‘₯ = 2β„Ž/(3 π‘π‘œπ‘‘β‘Ξ± ) ∴ Volume is maximum for π‘₯ = 2β„Ž/(3 π‘π‘œπ‘‘β‘Ξ± ) We need to find Height and Volume For Height Height of cylinder = 𝒉 – 𝒙 cot 𝜢 = β„Ž βˆ’ cot 𝛼 Γ— 2β„Ž/(3 π‘π‘œπ‘‘β‘γ€– 𝛼〗 ) = β„Ž βˆ’ 2β„Ž/3 = 𝒉/πŸ‘ Hence, Height of cylinder is one third of cone 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 = 4/27 ((πœ‹β„Ž^3)/cot^2⁑α ) V = πŸ’/πŸπŸ• 𝝅𝒉^πŸ‘.〖𝒕𝒂𝒏〗^𝟐⁑𝜢 Thus, greatest volume of cylinder is 4/27 πœ‹β„Ž^3 tan^2⁑𝛼

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.