Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 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 = 𝜋𝑟^2 ℎ V = 𝜋𝑟^2 ℎ 100 = 𝜋𝑟^2 ℎ 100/(𝜋𝑟^2 )=ℎ ℎ=(100/(𝜋𝑟^2 )) We need to minimize surface area We know that Surface area of Cylinder = 2π𝑟h + 2π𝑟2 S = 2𝜋𝑟ℎ + 2𝜋𝑟2 Putting value of h =(100/(𝜋𝑟^2 ℎ)) S = 2πr (100/(𝜋𝑟^2 ))+"2πr2 " S = 2(100/𝑟)+"2πr2 " S = 200/𝑟+2𝜋𝑟^2 S = 200r –1 + 2𝜋𝑟^2 We need to Minimize S S = 200r –1 + 2𝜋𝑟^2 Diff w.r.t.𝒓 𝑑𝑠/𝑑𝑟=𝑑(200"r –1 + " 2𝜋𝑟^2 )/𝑑𝑟 𝑑𝑠/𝑑𝑟=200(−1) 𝑟^(−1−1)+2𝜋 ×2𝑟^(2−1) 𝑑𝑠/𝑑𝑟=−200𝑟^(−2)+4𝜋𝑟 Putting 𝒅𝒔/𝒅𝒓=𝟎 −200𝑟^(−2)+4𝜋𝑟=0 (−200)/𝑟^2 +4𝜋𝑟=0 (−200+4𝜋𝑟^3)/𝑟^2 =0 −200+4𝜋𝑟^3=0 4𝜋𝑟^3=200 𝜋𝑟^3=200/4 〖𝜋𝑟〗^3=50 𝑟^3=(50/𝜋) 𝑟=(50/𝜋)^(1/3) Now, Finding (𝒅^𝟐 𝒔)/(𝒅𝒓^𝟐 ) (𝑑^2 𝑠)/(𝑑𝑟^2 )=−200𝑟^(−2)+4𝜋𝑟 Diff w.r.t 𝑟 𝑑𝑠/𝑑𝑟=𝑑(−200𝑟^(−2)+4𝜋𝑟)/𝑑𝑟 (𝑑^2 𝑠)/(𝑑𝑟^2 )=−200(−2) 𝑟^(−2−1)+4𝜋 (𝑑^2 𝑠)/(𝑑𝑟^2 )=400𝑟^(−3)+4𝜋 Putting value of 𝑟=(50/𝜋)^(1/3) ├ (𝑑^2 𝑠)/(𝑑𝑟^2 )┤|_(𝑟=(50/𝜋)^(1/3) )=400/((50/𝜋)^(1/3) )^3 +4𝜋=400/((50/𝜋) )+4𝜋 =600𝜋/50=12𝜋 > 0 (𝑑^2 𝑠)/(𝑑𝑟^2 )>0 at 𝑟=(50/𝜋)^(1/3) Hence S is minimum at 𝑟=(50/𝜋)^(1/3) Now, Finding ℎ ℎ = (100/(𝜋𝑟^2 )) Putting value of 𝑟=(50/𝜋)^(1/3) cm ℎ = 100/〖𝜋[(50/𝜋)^(1/3) ]〗^2 = 100/〖𝜋(50/𝜋)〗^(2/3) = 100/𝜋 (𝜋/50)^(2/3) = 100/𝜋 × 𝜋^(2/3)/50^(2/3) = (2 × 50)/50^(2/3) × 𝜋^(2/3)/𝜋 = (2 × 50^(1 − 2/3))/𝜋^(1 − 2/3) = 2 × 50^(1/3)/𝜋^(1/3) = 2(50/𝜋)^(1/3) Hence total surface area is least when Radius of base is (𝟓𝟎/𝜋)^(𝟏/𝟑) & 𝒉=𝟐 (𝟓𝟎/𝝅)^(𝟏/𝟑) cm.

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.