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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan –1 13 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 𝑟2+𝜋𝑟𝑙 S = π 𝑟2+𝜋𝑟𝑙 S – π 𝑟2=𝜋𝑟𝑙 𝑆 − 𝜋 𝑟2 𝜋𝑟=𝑙 𝑙 = 𝑆 − 𝜋 𝑟2 𝜋𝑟 We need to find minimize volume of a cone & show that semi vertical angle is sin −13 i.e. θ =𝑠𝑖𝑛 −13 sin θ = 13 We know that sin θ = 𝑟𝑙 Volume of a cone = 13𝜋 𝑟2ℎ V = 13𝜋 𝑟2 𝑙2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟2𝜋𝑟2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 𝜋2 𝑟2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 − 𝜋 𝑟2 𝑟2 𝜋2 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 − 𝜋2 𝑟4 𝜋2 𝑟2𝑟 V = 𝜋 𝑟23𝜋𝑟 𝑠−𝜋 𝑟22− 𝜋2 𝑟4 V = 𝑟3 𝑠2+ 𝜋 𝑟22−2𝑆 𝜋𝑟2− 𝜋2 𝑟4 V = 𝑟3 𝑠2+ 𝜋2 𝑟4−2𝑆𝜋 𝑟2− 𝜋2 𝑟4 V = 𝑟3 𝑠2−2 𝑆𝜋 𝑟2 V = 13 𝑟2 𝑠2−2 𝑠 𝜋 𝑟2 V = 13 𝑟2 𝑠2−2 𝑠𝜋 𝑟4 Diff w.r.t 𝑟 𝑑𝑉𝑑𝑟= 13 𝑑𝑑𝑟 𝑠2 𝑟2−2 𝑠𝜋 𝑟2 𝑑𝑣𝑑𝑟= 13× 12 𝑠2 𝑟2−2 𝑠𝜋 𝑟2× 2 𝑠2𝑟−2𝑠𝜋 𝑟3 𝑑𝑣𝑑𝑟= 1 𝑠2𝑟 − 4𝑠𝜋 𝑟33 𝑠2 𝑟2 − 2 𝑠𝜋 𝑟2 Putting 𝑑𝑣𝑑𝑟=0 13 𝑠2 𝑟2 − 2𝑠𝜋 𝑟4× 𝑠2𝑟−4𝜋𝑠 𝑟3=0 Hence, S = 4𝜋 𝑟2 Now, Surface area of cone = 𝜋 𝑟2+𝜋𝑟𝑙 𝑆=𝜋 𝑟2+𝜋𝑟𝑙 Putting S = 4𝜋 𝑟2 4𝜋 𝑟2=𝜋 𝑟2+𝜋𝑟𝑙 𝜋 𝑟2+𝜋𝑟𝑙=4𝜋 𝑟2 Dividing both sides by 𝜋𝑟 𝜋 𝑟2+ 𝜋𝑟𝑙𝜋𝑟= 4𝜋 𝑟2𝜋𝑟 𝑟+𝑙=4𝑟 𝑙=4𝑟−𝑟 𝑙=3𝑟 𝑙𝑟=3 𝑟𝑙= 13 But we know that sin θ = 𝑟𝑙 Putting value of 𝑟𝑙 sin θ = 13 θ = 𝒔𝒊𝒏−𝟏 𝟏𝟑 Hence proved

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.