Check sibling questions

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

Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

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 = 2𝜋𝑟^2+ 2𝜋𝑟ℎ S = 2𝜋𝑟^2+ 2𝜋𝑟ℎ S – 2𝜋𝑟^2= 2𝜋𝑟ℎ (𝑆 − 2𝜋𝑟^2)/2𝜋𝑟=ℎ ℎ=(𝑆 − 2𝜋𝑟^2)/2𝜋𝑟 Volume of Cylinder = 𝜋𝑟2ℎ V = 𝜋𝑟2ℎ We need to maximum volume Now, V = πr2h V = πr2 ((𝑆 − 2𝜋𝑟^2)/2𝜋𝑟) V = (𝜋𝑟^2)/2𝜋𝑟 (𝑆 −2𝜋𝑟^2 ) V = 𝑟/2 (𝑆 −2𝜋𝑟^2 ) V = 1/2 (𝑆𝑟 −2𝜋𝑟^3 ) Diff w.r.t 𝒓 𝑑𝑉/𝑑𝑟=1/2 𝑑(𝑆𝑟−2𝜋𝑟^3 )/𝑑𝑟 𝑑𝑉/𝑑𝑟=1/2 (𝑆−6𝜋𝑟^2 ) Putting 𝒅𝑽/𝒅𝒓=𝟎 1/2 (𝑆−6𝜋𝑟^2 )=0 𝑆−6𝜋𝑟^2=0 Putting value of 𝑆 = 2𝜋𝑟2+ 2𝜋𝑟ℎ (2𝜋𝑟^2+2𝜋𝑟ℎ)−6𝜋𝑟^2=0 −4𝜋 𝑟2 + 2𝜋𝑟ℎ = 0 2𝜋𝑟ℎ (−2𝑟+ℎ)=0 2𝜋𝑟ℎ(ℎ−2𝑟)=0 ℎ−2𝑟=0 ℎ=2𝑟 Finding (𝒅^𝟐 𝒗)/(𝒅𝒓^𝟐 ) 𝑑𝑉/𝑑𝑟=1/2 (𝑠−6𝜋𝑟^2 ) (𝑑^2 𝑉)/(𝑑𝑟^2 )=1/2 𝑑(𝑆 − 6𝜋𝑟^2 )/𝑑𝑟 \ (𝑑^2 𝑣)/(𝑑𝑟^2 )=1/2 (0−12𝜋𝑟) (𝑑^2 𝑣)/(𝑑𝑟^2 )=−6𝜋𝑟 ∴ (𝑑^2 𝑣)/(𝑑𝑟^2 )<0 for ℎ=2𝑟 Hence, Volume of a cylinder is Maximum when 𝒉=𝟐𝒓

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.