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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 π‘Ÿ , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & ΞΈ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = Ο€π‘Ÿ^2+πœ‹π‘Ÿπ‘™ S = Ο€π‘Ÿ^2+πœ‹π‘Ÿπ‘™ S – Ο€π‘Ÿ^2=πœ‹π‘Ÿπ‘™ (𝑆 βˆ’ πœ‹π‘Ÿ^2)/(\ πœ‹π‘Ÿ)=𝑙 𝑙 = (𝑆 βˆ’ πœ‹π‘Ÿ^2)/(\ πœ‹π‘Ÿ) We need to find minimize volume of a cone & show that semi vertical angle is sin (βˆ’1)/3 i.e. ΞΈ =𝑠𝑖𝑛 (βˆ’1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π‘Ÿ/𝑙 Volume of a cone = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘Ÿ^2 √(𝑙^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ)^2βˆ’π‘Ÿ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (βˆ’1)/3 i.e. ΞΈ =𝑠𝑖𝑛 (βˆ’1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π‘Ÿ/𝑙 Volume of a cone = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘Ÿ^2 √(𝑙^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ)^2βˆ’π‘Ÿ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (βˆ’1)/3 i.e. ΞΈ =𝑠𝑖𝑛 (βˆ’1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π‘Ÿ/𝑙 Volume of a cone = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘Ÿ^2 √(𝑙^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ)^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2/(πœ‹^2 π‘Ÿ^2 )βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2 βˆ’ πœ‹π‘Ÿ^2 (π‘Ÿ^2 ))/(πœ‹^2 π‘Ÿ^2 )) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2 βˆ’ πœ‹^2 π‘Ÿ^4)/(πœ‹^2 π‘Ÿ^2 π‘Ÿ)) V = (πœ‹π‘Ÿ^2)/3πœ‹π‘Ÿ √((π‘ βˆ’πœ‹π‘Ÿ^2 )^2βˆ’πœ‹^2 π‘Ÿ^4 ) V = ((π‘Ÿ))/3 √(γ€–(𝑠)^2+(πœ‹π‘Ÿ^2 )γ€—^2βˆ’2𝑆 γ€–πœ‹π‘Ÿγ€—^2βˆ’πœ‹^2 π‘Ÿ^4 ) V = π‘Ÿ/3 √(𝑠^2+πœ‹^2 π‘Ÿ^4βˆ’2π‘†πœ‹π‘Ÿ^2βˆ’πœ‹^2 π‘Ÿ^4 ) V = π‘Ÿ/3 √(𝑠^2βˆ’2 π‘†πœ‹π‘Ÿ^2 ) V = 1/3 √(π‘Ÿ^2 (𝑠^2βˆ’2 𝑠 πœ‹π‘Ÿ^2 ) ) V = 1/3 √(π‘Ÿ^2 𝑠^2βˆ’2 π‘ πœ‹π‘Ÿ^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (π‘Ÿ^2 𝑠^2βˆ’2 π‘ πœ‹π‘Ÿ^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=𝑑(1/3 (π‘Ÿ^2 𝑠^2 βˆ’ 2π‘ πœ‹π‘Ÿ^4 ))/π‘‘π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=1/3 [𝑠^2 (2π‘Ÿ)βˆ’2π‘ πœ‹ (4π‘Ÿ^3 )] 𝑑Z/π‘‘π‘Ÿ=1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] Putting 𝒅𝒁/𝒅𝒓 = 0 1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ]=0 2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3=0 2π‘Ÿπ‘ ^2=8π‘ πœ‹π‘Ÿ^3 (2𝑠^2)/(4π‘ πœ‹ )=π‘Ÿ^3/π‘Ÿ 𝑠/(4πœ‹ )=π‘Ÿ^2 𝑠=4πœ‹π‘Ÿ^2 Finding (𝒅^𝟐 𝐙)/(𝐝𝒓^𝟐 ) 𝑑Z/π‘‘π‘Ÿ=1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] Diff w.r.t π‘₯ (𝑑^2 Z)/(π‘‘π‘Ÿ^2 ) = 𝑑/π‘‘π‘Ÿ [1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] " " ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2βˆ’8π‘ πœ‹(3π‘Ÿ^2) ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2βˆ’24π‘ πœ‹π‘Ÿ^2 ] Putting 𝑠=4πœ‹π‘Ÿ^2 (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2γ€–(4πœ‹π‘Ÿ^2)γ€—^2βˆ’24(4πœ‹π‘Ÿ^2)πœ‹π‘Ÿ^2 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [32πœ‹^2 π‘Ÿ^4βˆ’96πœ‹^2 π‘Ÿ^4 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [βˆ’64πœ‹^2 π‘Ÿ^4 ] Since (𝑑^2 Z)/(π‘‘π‘Ÿ^2 ) < 0 for 𝑠=4πœ‹π‘Ÿ^2 Volume is maximum for 𝑠=4πœ‹π‘Ÿ^2 Now, Surface area of cone = πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ 𝑆=πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ Putting S = 4πœ‹π‘Ÿ^2 4πœ‹π‘Ÿ^2=πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™=4πœ‹π‘Ÿ^2 Dividing both sides by πœ‹π‘Ÿ (πœ‹π‘Ÿ^2+ πœ‹π‘Ÿπ‘™)/πœ‹π‘Ÿ=(4πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ π‘Ÿ+𝑙=4π‘Ÿ 𝑙=4π‘Ÿβˆ’π‘Ÿ 𝑙=3π‘Ÿ 𝑙/π‘Ÿ=3 π‘Ÿ/𝑙=1/3 But we know that sin ΞΈ =π‘Ÿ/𝑙 Putting value of π‘Ÿ/𝑙 sin ΞΈ =1/3 ΞΈ =γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑ 𝟏/πŸ‘ 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.