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

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

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.