Ex 6.5, 20 - Show that cylinder of given surface, max volume - Ex 6.5

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


Ex 6.5,20 Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base. Let 𝑟 , ℎ be the Radius & Height of Cylinder respectively & 𝑉 , 𝑆 be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2𝜋 𝑟﷮2﷯+ 2𝜋𝑟ℎ S = 2𝜋 𝑟﷮2﷯+ 2𝜋𝑟ℎ S – 2𝜋 𝑟﷮2﷯= 2𝜋𝑟ℎ 𝑆 − 2𝜋 𝑟﷮2﷯﷮2𝜋𝑟﷯=ℎ ℎ= 𝑆 − 2𝜋 𝑟﷮2﷯﷮2𝜋𝑟﷯ Volume of Cylinder = 𝜋𝑟2ℎ V = 𝜋𝑟2ℎ We need to maximum volume Now, V = πr2h V = πr2 𝑆 − 2𝜋 𝑟﷮2﷯﷮2𝜋𝑟﷯﷯ V = 𝜋 𝑟﷮2﷯﷮2𝜋𝑟﷯ 𝑆 −2𝜋 𝑟﷮2﷯﷯ V = 𝑟﷮2﷯ 𝑆 −2𝜋 𝑟﷮2﷯﷯ V = 1﷮2﷯ 𝑆𝑟 −2𝜋 𝑟﷮3﷯﷯ Diff w.r.t 𝑟 𝑑𝑉﷮𝑑𝑟﷯= 1﷮2﷯ 𝑑 𝑠𝑟−2𝜋 𝑟﷮3﷯﷯﷮𝑑𝑟﷯ 𝑑𝑉﷮𝑑𝑟﷯= 1﷮2﷯ 𝑠−6𝜋 𝑟﷮2﷯﷯ Putting 𝑑𝑣﷮𝑑𝑟﷯=0 1﷮2﷯ 𝑠−6𝜋 𝑟﷮2﷯﷯=0 𝑠−6𝜋 𝑟﷮2﷯=0 Putting value of 𝑆 = 2𝜋𝑟2+ 2𝜋𝑟ℎ 2𝜋 𝑟﷮2﷯+2𝜋𝑟ℎ﷯−6𝜋 𝑟﷮2﷯=0 −4𝜋 𝑟2 + 2𝜋𝑟ℎ = 0 2𝜋𝑟ℎ −2𝑟+ℎ﷯=0 2𝜋𝑟 ℎ−2𝑟﷯=0 ℎ−2𝑟=0 ℎ=2𝑟 Finding 𝑑﷮2﷯𝑣﷮𝑑 𝑟﷮2﷯﷯ 𝑑𝑉﷮𝑑𝑟﷯= 1﷮2﷯ 𝑠−6𝜋 𝑟﷮2﷯﷯ 𝑑﷮2﷯𝑉﷮𝑑 𝑟﷮2﷯﷯= 1﷮2﷯ 𝑑 𝑆 − 6𝜋 𝑟﷮2﷯﷯﷮𝑑𝑟﷯ 𝑑﷮2﷯𝑣﷮𝑑 𝑟﷮2﷯﷯= 1﷮2﷯ 0−12𝜋𝑟﷯ 𝑑﷮2﷯𝑣﷮𝑑 𝑟﷮2﷯﷯=−6𝜋𝑟 < 0 ∴ 𝑑﷮2﷯𝑣﷮𝑑 𝑟﷮2﷯﷯<0 for ℎ=2𝑟 Hence volume of a cylinder is Maximum when 𝒉=𝟐𝒓

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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.