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Ex 6.5, 21 - Of all closed cylindrical cans of a given volume

Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives - Part 8

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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 = πœ‹π‘Ÿ^2 β„Ž V = πœ‹π‘Ÿ^2 β„Ž 100 = πœ‹π‘Ÿ^2 β„Ž 100/(πœ‹π‘Ÿ^2 )=β„Ž β„Ž=(100/(πœ‹π‘Ÿ^2 )) We need to minimize surface area We know that Surface area of Cylinder = 2Ο€π‘Ÿh + 2Ο€π‘Ÿ2 S = 2πœ‹π‘Ÿβ„Ž + 2πœ‹π‘Ÿ2 Putting value of h =(100/(πœ‹π‘Ÿ^2 β„Ž)) S = 2Ο€r (100/(πœ‹π‘Ÿ^2 ))+"2Ο€r2 " S = 2(100/π‘Ÿ)+"2Ο€r2 " S = 200/π‘Ÿ+2πœ‹π‘Ÿ^2 S = 200r –1 + 2πœ‹π‘Ÿ^2 We need to Minimize S S = 200r –1 + 2πœ‹π‘Ÿ^2 Diff w.r.t.𝒓 𝑑𝑠/π‘‘π‘Ÿ=𝑑(200"r –1 + " 2πœ‹π‘Ÿ^2 )/π‘‘π‘Ÿ 𝑑𝑠/π‘‘π‘Ÿ=200(βˆ’1) π‘Ÿ^(βˆ’1βˆ’1)+2πœ‹ Γ—2π‘Ÿ^(2βˆ’1) 𝑑𝑠/π‘‘π‘Ÿ=βˆ’200π‘Ÿ^(βˆ’2)+4πœ‹π‘Ÿ Putting 𝒅𝒔/𝒅𝒓=𝟎 βˆ’200π‘Ÿ^(βˆ’2)+4πœ‹π‘Ÿ=0 (βˆ’200)/π‘Ÿ^2 +4πœ‹π‘Ÿ=0 (βˆ’200+4πœ‹π‘Ÿ^3)/π‘Ÿ^2 =0 βˆ’200+4πœ‹π‘Ÿ^3=0 4πœ‹π‘Ÿ^3=200 πœ‹π‘Ÿ^3=200/4 γ€–πœ‹π‘Ÿγ€—^3=50 π‘Ÿ^3=(50/πœ‹) π‘Ÿ=(50/πœ‹)^(1/3) Now, Finding (𝒅^𝟐 𝒔)/(𝒅𝒓^𝟐 ) (𝑑^2 𝑠)/(π‘‘π‘Ÿ^2 )=βˆ’200π‘Ÿ^(βˆ’2)+4πœ‹π‘Ÿ Diff w.r.t π‘Ÿ 𝑑𝑠/π‘‘π‘Ÿ=𝑑(βˆ’200π‘Ÿ^(βˆ’2)+4πœ‹π‘Ÿ)/π‘‘π‘Ÿ (𝑑^2 𝑠)/(π‘‘π‘Ÿ^2 )=βˆ’200(βˆ’2) π‘Ÿ^(βˆ’2βˆ’1)+4πœ‹ (𝑑^2 𝑠)/(π‘‘π‘Ÿ^2 )=400π‘Ÿ^(βˆ’3)+4πœ‹ Putting value of π‘Ÿ=(50/πœ‹)^(1/3) β”œ (𝑑^2 𝑠)/(π‘‘π‘Ÿ^2 )─|_(π‘Ÿ=(50/πœ‹)^(1/3) )=400/((50/πœ‹)^(1/3) )^3 +4πœ‹=400/((50/πœ‹) )+4πœ‹ =600πœ‹/50=12πœ‹ > 0 (𝑑^2 𝑠)/(π‘‘π‘Ÿ^2 )>0 at π‘Ÿ=(50/πœ‹)^(1/3) Hence S is minimum at π‘Ÿ=(50/πœ‹)^(1/3) Now, Finding β„Ž β„Ž = (100/(πœ‹π‘Ÿ^2 )) Putting value of π‘Ÿ=(50/πœ‹)^(1/3) cm β„Ž = 100/γ€–πœ‹[(50/πœ‹)^(1/3) ]γ€—^2 = 100/γ€–πœ‹(50/πœ‹)γ€—^(2/3) = 100/πœ‹ (πœ‹/50)^(2/3) = 100/πœ‹ Γ— πœ‹^(2/3)/50^(2/3) = (2 Γ— 50)/50^(2/3) Γ— πœ‹^(2/3)/πœ‹ = (2 Γ— 50^(1 βˆ’ 2/3))/πœ‹^(1 βˆ’ 2/3) = 2 Γ— 50^(1/3)/πœ‹^(1/3) = 2(50/πœ‹)^(1/3) Hence total surface area is least when Radius of base is (πŸ“πŸŽ/πœ‹)^(𝟏/πŸ‘) & 𝒉=𝟐 (πŸ“πŸŽ/𝝅)^(𝟏/πŸ‘) cm.

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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.