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

Transcript

Ex 6.5, 21 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area? Let r ,& h be the radius & height of cylinder respectively & V & S be the volume & surface area of cylinder respectively Given volume = 100 cm3 We know that Volume of a Cylinder = ๐œ‹๐‘Ÿ^2 โ„Ž V = ๐œ‹๐‘Ÿ^2 โ„Ž 100 = ๐œ‹๐‘Ÿ^2 โ„Ž 100/(๐œ‹๐‘Ÿ^2 )=โ„Ž โ„Ž=(100/(๐œ‹๐‘Ÿ^2 )) We need to minimize surface area We know that Surface area of Cylinder = 2ฯ€๐‘Ÿh + 2ฯ€๐‘Ÿ2 S = 2๐œ‹๐‘Ÿโ„Ž + 2๐œ‹๐‘Ÿ2 Putting value of h =(100/(๐œ‹๐‘Ÿ^2 โ„Ž)) S = 2ฯ€r (100/(๐œ‹๐‘Ÿ^2 ))+"2ฯ€r2 " S = 2(100/๐‘Ÿ)+"2ฯ€r2 " S = 200/๐‘Ÿ+2๐œ‹๐‘Ÿ^2 S = 200r โ€“1 + 2๐œ‹๐‘Ÿ^2 We need to Minimize S S = 200r โ€“1 + 2๐œ‹๐‘Ÿ^2 Diff w.r.t.๐’“ ๐‘‘๐‘ /๐‘‘๐‘Ÿ=๐‘‘(200"r โ€“1 + " 2๐œ‹๐‘Ÿ^2 )/๐‘‘๐‘Ÿ ๐‘‘๐‘ /๐‘‘๐‘Ÿ=200(โˆ’1) ๐‘Ÿ^(โˆ’1โˆ’1)+2๐œ‹ ร—2๐‘Ÿ^(2โˆ’1) ๐‘‘๐‘ /๐‘‘๐‘Ÿ=โˆ’200๐‘Ÿ^(โˆ’2)+4๐œ‹๐‘Ÿ Putting ๐’…๐’”/๐’…๐’“=๐ŸŽ โˆ’200๐‘Ÿ^(โˆ’2)+4๐œ‹๐‘Ÿ=0 (โˆ’200)/๐‘Ÿ^2 +4๐œ‹๐‘Ÿ=0 (โˆ’200+4๐œ‹๐‘Ÿ^3)/๐‘Ÿ^2 =0 โˆ’200+4๐œ‹๐‘Ÿ^3=0 4๐œ‹๐‘Ÿ^3=200 ๐œ‹๐‘Ÿ^3=200/4 ใ€–๐œ‹๐‘Ÿใ€—^3=50 ๐‘Ÿ^3=(50/๐œ‹) ๐‘Ÿ=(50/๐œ‹)^(1/3) Now, Finding (๐’…^๐Ÿ ๐’”)/(๐’…๐’“^๐Ÿ ) (๐‘‘^2 ๐‘ )/(๐‘‘๐‘Ÿ^2 )=โˆ’200๐‘Ÿ^(โˆ’2)+4๐œ‹๐‘Ÿ Diff w.r.t ๐‘Ÿ ๐‘‘๐‘ /๐‘‘๐‘Ÿ=๐‘‘(โˆ’200๐‘Ÿ^(โˆ’2)+4๐œ‹๐‘Ÿ)/๐‘‘๐‘Ÿ (๐‘‘^2 ๐‘ )/(๐‘‘๐‘Ÿ^2 )=โˆ’200(โˆ’2) ๐‘Ÿ^(โˆ’2โˆ’1)+4๐œ‹ (๐‘‘^2 ๐‘ )/(๐‘‘๐‘Ÿ^2 )=400๐‘Ÿ^(โˆ’3)+4๐œ‹ Putting value of ๐‘Ÿ=(50/๐œ‹)^(1/3) โ”œ (๐‘‘^2 ๐‘ )/(๐‘‘๐‘Ÿ^2 )โ”ค|_(๐‘Ÿ=(50/๐œ‹)^(1/3) )=400/((50/๐œ‹)^(1/3) )^3 +4๐œ‹=400/((50/๐œ‹) )+4๐œ‹ =600๐œ‹/50=12๐œ‹ > 0 (๐‘‘^2 ๐‘ )/(๐‘‘๐‘Ÿ^2 )>0 at ๐‘Ÿ=(50/๐œ‹)^(1/3) Hence S is minimum at ๐‘Ÿ=(50/๐œ‹)^(1/3) Now, Finding โ„Ž โ„Ž = (100/(๐œ‹๐‘Ÿ^2 )) Putting value of ๐‘Ÿ=(50/๐œ‹)^(1/3) cm โ„Ž = 100/ใ€–๐œ‹[(50/๐œ‹)^(1/3) ]ใ€—^2 = 100/ใ€–๐œ‹(50/๐œ‹)ใ€—^(2/3) = 100/๐œ‹ (๐œ‹/50)^(2/3) = 100/๐œ‹ ร— ๐œ‹^(2/3)/ใ€–50ใ€—^(2/3) = (2 ร— 50)/ใ€–50ใ€—^(2/3) ร— ๐œ‹^(2/3)/๐œ‹ = (2 ร— ใ€–50ใ€—^(1 โˆ’ 2/3))/๐œ‹^(1 โˆ’ 2/3) = 2 ร— ใ€–50ใ€—^(1/3)/๐œ‹^(1/3) = 2(50/๐œ‹)^(1/3) Hence total surface area is least when Radius of base is (๐Ÿ“๐ŸŽ/๐œ‹)^(๐Ÿ/๐Ÿ‘) & ๐’‰=๐Ÿ (๐Ÿ“๐ŸŽ/๐…)^(๐Ÿ/๐Ÿ‘) cm.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.