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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

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 ๐‘Ÿ, โ„Ž be the Radius & Height of Cylinder respectively & ๐‘‰, ๐‘† be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2๐œ‹๐‘Ÿ^2+ 2๐œ‹๐‘Ÿโ„Ž S = 2๐œ‹๐‘Ÿ^2+ 2๐œ‹๐‘Ÿโ„Ž S โ€“ 2๐œ‹๐‘Ÿ^2= 2๐œ‹๐‘Ÿโ„Ž (๐‘† โˆ’ 2๐œ‹๐‘Ÿ^2)/2๐œ‹๐‘Ÿ=โ„Ž โ„Ž=(๐‘† โˆ’ 2๐œ‹๐‘Ÿ^2)/2๐œ‹๐‘Ÿ Volume of Cylinder = ๐œ‹๐‘Ÿ2โ„Ž V = ๐œ‹๐‘Ÿ2โ„Ž We need to maximum volume Now, V = ฯ€r2h V = ฯ€r2 ((๐‘† โˆ’ 2๐œ‹๐‘Ÿ^2)/2๐œ‹๐‘Ÿ) V = (๐œ‹๐‘Ÿ^2)/2๐œ‹๐‘Ÿ (๐‘† โˆ’2๐œ‹๐‘Ÿ^2 ) V = ๐‘Ÿ/2 (๐‘† โˆ’2๐œ‹๐‘Ÿ^2 ) V = 1/2 (๐‘†๐‘Ÿ โˆ’2๐œ‹๐‘Ÿ^3 ) Diff w.r.t ๐’“ ๐‘‘๐‘‰/๐‘‘๐‘Ÿ=1/2 ๐‘‘(๐‘†๐‘Ÿโˆ’2๐œ‹๐‘Ÿ^3 )/๐‘‘๐‘Ÿ ๐‘‘๐‘‰/๐‘‘๐‘Ÿ=1/2 (๐‘†โˆ’6๐œ‹๐‘Ÿ^2 ) Putting ๐’…๐‘ฝ/๐’…๐’“=๐ŸŽ 1/2 (๐‘†โˆ’6๐œ‹๐‘Ÿ^2 )=0 ๐‘†โˆ’6๐œ‹๐‘Ÿ^2=0 Putting value of ๐‘† = 2๐œ‹๐‘Ÿ2+ 2๐œ‹๐‘Ÿโ„Ž (2๐œ‹๐‘Ÿ^2+2๐œ‹๐‘Ÿโ„Ž)โˆ’6๐œ‹๐‘Ÿ^2=0 โˆ’4๐œ‹ ๐‘Ÿ2 + 2๐œ‹๐‘Ÿโ„Ž = 0 2๐œ‹๐‘Ÿโ„Ž (โˆ’2๐‘Ÿ+โ„Ž)=0 2๐œ‹๐‘Ÿโ„Ž(โ„Žโˆ’2๐‘Ÿ)=0 โ„Žโˆ’2๐‘Ÿ=0 โ„Ž=2๐‘Ÿ Finding (๐’…^๐Ÿ ๐’—)/(๐’…๐’“^๐Ÿ ) ๐‘‘๐‘‰/๐‘‘๐‘Ÿ=1/2 (๐‘ โˆ’6๐œ‹๐‘Ÿ^2 ) (๐‘‘^2 ๐‘‰)/(๐‘‘๐‘Ÿ^2 )=1/2 ๐‘‘(๐‘† โˆ’ 6๐œ‹๐‘Ÿ^2 )/๐‘‘๐‘Ÿ \ (๐‘‘^2 ๐‘ฃ)/(๐‘‘๐‘Ÿ^2 )=1/2 (0โˆ’12๐œ‹๐‘Ÿ) (๐‘‘^2 ๐‘ฃ)/(๐‘‘๐‘Ÿ^2 )=โˆ’6๐œ‹๐‘Ÿ โˆด (๐‘‘^2 ๐‘ฃ)/(๐‘‘๐‘Ÿ^2 )<0 for โ„Ž=2๐‘Ÿ Hence, Volume of a cylinder is Maximum when ๐’‰=๐Ÿ๐’“

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.