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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 6.3, 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 = π𝑟^2+𝜋𝑟𝑙 S = π𝑟^2+𝜋𝑟𝑙 S – π𝑟^2=𝜋𝑟𝑙 (𝑆 − 𝜋𝑟^2)/(\ 𝜋𝑟)=𝑙 𝑙 = (𝑆 − 𝜋𝑟^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 𝜋𝑟^2 ℎ V = 1/3 𝜋𝑟^2 √(𝑙^2−𝑟^2 ) V = 1/3 𝜋𝑟^2 √(((𝑠 − 𝜋𝑟^2)/𝜋𝑟)^2−𝑟^2 ) 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 𝜋𝑟^2 ℎ V = 1/3 𝜋𝑟^2 √(𝑙^2−𝑟^2 ) V = 1/3 𝜋𝑟^2 √(((𝑠 − 𝜋𝑟^2)/𝜋𝑟)^2−𝑟^2 ) 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 𝜋𝑟^2 ℎ V = 1/3 𝜋𝑟^2 √(𝑙^2−𝑟^2 ) V = 1/3 𝜋𝑟^2 √(((𝑠 − 𝜋𝑟^2)/𝜋𝑟)^2−𝑟^2 ) V = 1/3 𝜋𝑟^2 √((𝑠 − 𝜋𝑟^2 )^2/(𝜋^2 𝑟^2 )−𝑟^2 ) V = 1/3 𝜋𝑟^2 √(((𝑠 − 𝜋𝑟^2 )^2 − 𝜋𝑟^2 (𝑟^2 ))/(𝜋^2 𝑟^2 )) V = 1/3 𝜋𝑟^2 √(((𝑠 − 𝜋𝑟^2 )^2 − 𝜋^2 𝑟^4)/(𝜋^2 𝑟^2 𝑟)) V = (𝜋𝑟^2)/3𝜋𝑟 √((𝑠−𝜋𝑟^2 )^2−𝜋^2 𝑟^4 ) V = ((𝑟))/3 √(〖(𝑠)^2+(𝜋𝑟^2 )〗^2−2𝑆 〖𝜋𝑟〗^2−𝜋^2 𝑟^4 ) V = 𝑟/3 √(𝑠^2+𝜋^2 𝑟^4−2𝑆𝜋𝑟^2−𝜋^2 𝑟^4 ) V = 𝑟/3 √(𝑠^2−2 𝑆𝜋𝑟^2 ) V = 1/3 √(𝑟^2 (𝑠^2−2 𝑠 𝜋𝑟^2 ) ) V = 1/3 √(𝑟^2 𝑠^2−2 𝑠𝜋𝑟^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (𝑟^2 𝑠^2−2 𝑠𝜋𝑟^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t 𝑟 𝑑Z/𝑑𝑟=𝑑(1/3 (𝑟^2 𝑠^2 − 2𝑠𝜋𝑟^4 ))/𝑑𝑟 𝑑Z/𝑑𝑟=1/3 [𝑠^2 (2𝑟)−2𝑠𝜋 (4𝑟^3 )] 𝑑Z/𝑑𝑟=1/3 [2𝑟𝑠^2−8𝑠𝜋𝑟^3 ] Putting 𝒅𝒁/𝒅𝒓 = 0 1/3 [2𝑟𝑠^2−8𝑠𝜋𝑟^3 ]=0 2𝑟𝑠^2−8𝑠𝜋𝑟^3=0 2𝑟𝑠^2=8𝑠𝜋𝑟^3 (2𝑠^2)/(4𝑠𝜋 )=𝑟^3/𝑟 𝑠/(4𝜋 )=𝑟^2 𝑠=4𝜋𝑟^2 Finding (𝒅^𝟐 𝐙)/(𝐝𝒓^𝟐 ) 𝑑Z/𝑑𝑟=1/3 [2𝑟𝑠^2−8𝑠𝜋𝑟^3 ] Diff w.r.t 𝑥 (𝑑^2 Z)/(𝑑𝑟^2 ) = 𝑑/𝑑𝑟 [1/3 [2𝑟𝑠^2−8𝑠𝜋𝑟^3 ] " " ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2−8𝑠𝜋(3𝑟^2) ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2−24𝑠𝜋𝑟^2 ] Putting 𝑠=4𝜋𝑟^2 (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2〖(4𝜋𝑟^2)〗^2−24(4𝜋𝑟^2)𝜋𝑟^2 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [32𝜋^2 𝑟^4−96𝜋^2 𝑟^4 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [−64𝜋^2 𝑟^4 ] Since (𝑑^2 Z)/(𝑑𝑟^2 ) < 0 for 𝑠=4𝜋𝑟^2 Volume is maximum for 𝑠=4𝜋𝑟^2 Now, Surface area of cone = 𝜋𝑟^2+𝜋𝑟𝑙 𝑆=𝜋𝑟^2+𝜋𝑟𝑙 Putting S = 4𝜋𝑟^2 4𝜋𝑟^2=𝜋𝑟^2+𝜋𝑟𝑙 𝜋𝑟^2+𝜋𝑟𝑙=4𝜋𝑟^2 Dividing both sides by 𝜋𝑟 (𝜋𝑟^2+ 𝜋𝑟𝑙)/𝜋𝑟=(4𝜋𝑟^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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.