Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives (Term 1)

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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