Ex 6.5,26 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan –1 1﷮3﷯﷯ Let 𝑟 , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & θ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = π 𝑟﷮2﷯+𝜋𝑟𝑙 S = π 𝑟﷮2﷯+𝜋𝑟𝑙 S – π 𝑟﷮2﷯=𝜋𝑟𝑙 𝑆 − 𝜋 𝑟﷮2﷯﷮ 𝜋𝑟﷯=𝑙 𝑙 = 𝑆 − 𝜋 𝑟﷮2﷯﷮ 𝜋𝑟﷯ We need to find minimize volume of a cone & show that semi vertical angle is sin −1﷮3﷯ i.e. θ =𝑠𝑖𝑛 −1﷮3﷯ sin θ = 1﷮3﷯ We know that sin θ = 𝑟﷮𝑙﷯ Volume of a cone = 1﷮3﷯𝜋 𝑟﷮2﷯ℎ V = 1﷮3﷯𝜋 𝑟﷮2﷯ ﷮ 𝑙﷮2﷯− 𝑟﷮2﷯﷯ V = 1﷮3﷯𝜋 𝑟﷮2﷯ ﷮ 𝑠 − 𝜋 𝑟﷮2﷯﷮𝜋𝑟﷯﷯﷮2﷯− 𝑟﷮2﷯﷯ V = 1﷮3﷯𝜋 𝑟﷮2﷯ ﷮ 𝑠 − 𝜋 𝑟﷮2﷯﷯﷮2﷯﷮ 𝜋﷮2﷯ 𝑟﷮2﷯﷯− 𝑟﷮2﷯﷯ V = 1﷮3﷯𝜋 𝑟﷮2﷯ ﷮ 𝑠 − 𝜋 𝑟﷮2﷯﷯﷮2﷯ − 𝜋 𝑟﷮2﷯ 𝑟﷮2﷯﷯﷮ 𝜋﷮2﷯ 𝑟﷮2﷯﷯﷯ V = 1﷮3﷯𝜋 𝑟﷮2﷯ ﷮ 𝑠 − 𝜋 𝑟﷮2﷯﷯﷮2﷯ − 𝜋﷮2﷯ 𝑟﷮4﷯﷮ 𝜋﷮2﷯ 𝑟﷮2﷯𝑟﷯﷯ V = 𝜋 𝑟﷮2﷯﷮3𝜋𝑟﷯ ﷮ 𝑠−𝜋 𝑟﷮2﷯﷯﷮2﷯− 𝜋﷮2﷯ 𝑟﷮4﷯﷯ V = 𝑟﷯﷮3﷯ ﷮ 𝑠﷯﷮2﷯+ 𝜋 𝑟﷮2﷯﷯﷮2﷯−2𝑆 𝜋𝑟﷮2﷯− 𝜋﷮2﷯ 𝑟﷮4﷯﷯ V = 𝑟﷮3﷯ ﷮ 𝑠﷮2﷯+ 𝜋﷮2﷯ 𝑟﷮4﷯−2𝑆𝜋 𝑟﷮2﷯− 𝜋﷮2﷯ 𝑟﷮4﷯﷯ V = 𝑟﷮3﷯ ﷮ 𝑠﷮2﷯−2 𝑆𝜋 𝑟﷮2﷯﷯ V = 1﷮3﷯ ﷮ 𝑟﷮2﷯ 𝑠﷮2﷯−2 𝑠 𝜋 𝑟﷮2﷯﷯﷯ V = 1﷮3﷯ ﷮ 𝑟﷮2﷯ 𝑠﷮2﷯−2 𝑠𝜋 𝑟﷮4﷯﷯ Diff w.r.t 𝑟 𝑑𝑉﷮𝑑𝑟﷯= 1﷮3﷯ 𝑑﷮𝑑𝑟﷯ ﷮ 𝑠﷮2﷯ 𝑟﷮2﷯−2 𝑠𝜋 𝑟﷮2﷯﷯ ﷯ 𝑑𝑣﷮𝑑𝑟﷯= 1﷮3﷯× 1﷮2﷯ ﷮ 𝑠﷮2﷯ 𝑟﷮2﷯−2 𝑠𝜋 𝑟﷮2﷯﷯× 2 𝑠﷮2﷯𝑟−2𝑠𝜋 𝑟﷮3﷯﷯ 𝑑𝑣﷮𝑑𝑟﷯= 1 𝑠﷮2﷯𝑟 − 4𝑠𝜋 𝑟﷮3﷯﷮3 ﷮ 𝑠﷮2﷯ 𝑟﷮2﷯ − 2 𝑠𝜋 𝑟﷮2﷯﷯﷯ Putting 𝑑𝑣﷮𝑑𝑟﷯=0 1﷮3 ﷮ 𝑠﷮2﷯ 𝑟﷮2﷯ − 2𝑠𝜋 𝑟﷮4﷯﷯﷯× 𝑠﷮2﷯𝑟−4𝜋𝑠 𝑟﷮3﷯﷯=0 Hence, S = 4𝜋 𝑟﷮2﷯ Now, Surface area of cone = 𝜋 𝑟﷮2﷯+𝜋𝑟𝑙 𝑆=𝜋 𝑟﷮2﷯+𝜋𝑟𝑙 Putting S = 4𝜋 𝑟﷮2﷯ 4𝜋 𝑟﷮2﷯=𝜋 𝑟﷮2﷯+𝜋𝑟𝑙 𝜋 𝑟﷮2﷯+𝜋𝑟𝑙=4𝜋 𝑟﷮2﷯ Dividing both sides by 𝜋𝑟 𝜋 𝑟﷮2﷯+ 𝜋𝑟𝑙﷮𝜋𝑟﷯= 4𝜋 𝑟﷮2﷯﷮𝜋𝑟﷯ 𝑟+𝑙=4𝑟 𝑙=4𝑟−𝑟 𝑙=3𝑟 𝑙﷮𝑟﷯=3 𝑟﷮𝑙﷯= 1﷮3﷯ But we know that sin θ = 𝑟﷮𝑙﷯ Putting value of 𝑟﷮𝑙﷯ sin θ = 1﷮3﷯ θ = 𝒔𝒊𝒏﷮−𝟏﷯﷮ ﷯ 𝟏﷮𝟑﷯ Hence proved