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Ex 6.5, 20 - Show that cylinder of given surface and maximum volume

Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

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 π‘Ÿ, β„Ž be the Radius & Height of Cylinder respectively & 𝑉, 𝑆 be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2πœ‹π‘Ÿ^2+ 2πœ‹π‘Ÿβ„Ž S = 2πœ‹π‘Ÿ^2+ 2πœ‹π‘Ÿβ„Ž S – 2πœ‹π‘Ÿ^2= 2πœ‹π‘Ÿβ„Ž (𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ=β„Ž β„Ž=(𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ Volume of Cylinder = πœ‹π‘Ÿ2β„Ž V = πœ‹π‘Ÿ2β„Ž We need to maximum volume Now, V = Ο€r2h V = Ο€r2 ((𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ) V = (πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ (𝑆 βˆ’2πœ‹π‘Ÿ^2 ) V = π‘Ÿ/2 (𝑆 βˆ’2πœ‹π‘Ÿ^2 ) V = 1/2 (π‘†π‘Ÿ βˆ’2πœ‹π‘Ÿ^3 ) Diff w.r.t 𝒓 𝑑𝑉/π‘‘π‘Ÿ=1/2 𝑑(π‘†π‘Ÿβˆ’2πœ‹π‘Ÿ^3 )/π‘‘π‘Ÿ 𝑑𝑉/π‘‘π‘Ÿ=1/2 (π‘†βˆ’6πœ‹π‘Ÿ^2 ) Putting 𝒅𝑽/𝒅𝒓=𝟎 1/2 (π‘†βˆ’6πœ‹π‘Ÿ^2 )=0 π‘†βˆ’6πœ‹π‘Ÿ^2=0 Putting value of 𝑆 = 2πœ‹π‘Ÿ2+ 2πœ‹π‘Ÿβ„Ž (2πœ‹π‘Ÿ^2+2πœ‹π‘Ÿβ„Ž)βˆ’6πœ‹π‘Ÿ^2=0 βˆ’4πœ‹ π‘Ÿ2 + 2πœ‹π‘Ÿβ„Ž = 0 2πœ‹π‘Ÿβ„Ž (βˆ’2π‘Ÿ+β„Ž)=0 2πœ‹π‘Ÿβ„Ž(β„Žβˆ’2π‘Ÿ)=0 β„Žβˆ’2π‘Ÿ=0 β„Ž=2π‘Ÿ Finding (𝒅^𝟐 𝒗)/(𝒅𝒓^𝟐 ) 𝑑𝑉/π‘‘π‘Ÿ=1/2 (π‘ βˆ’6πœ‹π‘Ÿ^2 ) (𝑑^2 𝑉)/(π‘‘π‘Ÿ^2 )=1/2 𝑑(𝑆 βˆ’ 6πœ‹π‘Ÿ^2 )/π‘‘π‘Ÿ \ (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )=1/2 (0βˆ’12πœ‹π‘Ÿ) (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )=βˆ’6πœ‹π‘Ÿ ∴ (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )<0 for β„Ž=2π‘Ÿ Hence, Volume of a cylinder is Maximum when 𝒉=πŸπ’“

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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.