# Ex 6.5,21

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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

Serial order wise

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