# Ex 6.5,21 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 = πr2h V = 𝜋 𝑟2ℎ 100 = 𝜋 𝑟2ℎ 100𝜋 𝑟2=ℎ ℎ= 100𝜋 𝑟2 We need to minimize surface area We know that Surface area of Cylinder = 2π𝑟h + 2π𝑟2 S = 2𝜋𝑟ℎ + 2𝜋𝑟2 Putting value of h = 100𝜋 𝑟2ℎ S = 2πr 100𝜋 𝑟2+2πr2 S = 2 100𝑟+2πr2 S = 200𝑟+2𝜋 𝑟2 S = 200r –1 + 2𝜋 𝑟2 We need to Minimize S S = 200r –1 + 2𝜋 𝑟2 Diff w.r.t 𝑟 𝑑𝑠𝑑𝑟= 𝑑 200r –1 + 2𝜋 𝑟2𝑑𝑟 𝑑𝑠𝑑𝑟=200 −1 𝑟−1−1+2𝜋 ×2 𝑟2−1 𝑑𝑠𝑑𝑟=−200 𝑟−2+4𝜋𝑟 Putting 𝑑𝑠𝑑𝑟=0 −200 𝑟−2+4𝜋𝑟=0 −200 𝑟2+4𝜋𝑟=0 −200+4𝜋 𝑟3 𝑟2=0 −200+4𝜋 𝑟3=0 4𝜋 𝑟3=200 𝜋 𝑟3= 2004 𝜋𝑟3=50 𝑟3= 50𝜋 𝑟= 50𝜋 13 Now, Finding 𝑑2𝑠𝑑 𝑟2 𝑑2𝑠𝑑 𝑟2=−200 𝑟−2+4𝜋𝑟 Diff w.r.t 𝑟 𝑑𝑠𝑑𝑟= 𝑑 −200 𝑟−2+4𝜋𝑟𝑑𝑟 𝑑2𝑠𝑑 𝑟2=−200 −2 𝑟−2−1+4𝜋 𝑑2𝑠𝑑 𝑟2=400 𝑟−3+4𝜋 Putting value of 𝑟= 50𝜋 13 𝑑2𝑠𝑑 𝑟2𝑟= 50𝜋 13= 400 50𝜋 133+4𝜋= 400 50𝜋+4𝜋 = 600𝜋50=12𝜋 > 0 𝑑2𝑠𝑑 𝑟2>0 at 𝑟= 50𝜋 13 Hence S is minimum at 𝑟= 50𝜋 13 Now, Finding ℎ ℎ = 100𝜋 𝑟2 Putting value of 𝑟= 50𝜋 13 cm ℎ = 100 𝜋 50𝜋 132 = 100 𝜋 50𝜋 23 = 100𝜋 𝜋50 23 = 100𝜋 × 𝜋 23 50 23 = 2 × 50 50 23 × 𝜋 23𝜋 = 2 × 501 − 23 𝜋1 − 23 = 2 × 50 13 𝜋 13 = 2 50𝜋 13 Hence total surface area is least when Radius of base is 𝝅𝟓𝟎 𝟐𝟑 & 𝒉=𝟐 𝟓𝟎𝝅 𝟏𝟑 cm.

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Ex 6.5,21 You are here

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.