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

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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 = πr2h V = 𝜋 𝑟﷮2﷯ℎ 100 = 𝜋 𝑟﷮2﷯ℎ 100﷮𝜋 𝑟﷮2﷯﷯=ℎ ℎ= 100﷮𝜋 𝑟﷮2﷯﷯﷯ We need to minimize surface area We know that Surface area of Cylinder = 2π𝑟h + 2π𝑟2 S = 2𝜋𝑟ℎ + 2𝜋𝑟2 Putting value of h = 100﷮𝜋 𝑟﷮2﷯ℎ﷯﷯ S = 2πr 100﷮𝜋 𝑟﷮2﷯﷯﷯+2πr2 S = 2 100﷮𝑟﷯﷯+2πr2 S = 200﷮𝑟﷯+2𝜋 𝑟﷮2﷯ S = 200r –1 + 2𝜋 𝑟﷮2﷯ We need to Minimize S S = 200r –1 + 2𝜋 𝑟﷮2﷯ Diff w.r.t 𝑟 𝑑𝑠﷮𝑑𝑟﷯= 𝑑 200r –1 + 2𝜋 𝑟﷮2﷯﷯﷮𝑑𝑟﷯ 𝑑𝑠﷮𝑑𝑟﷯=200 −1﷯ 𝑟﷮−1−1﷯+2𝜋 ×2 𝑟﷮2−1﷯ 𝑑𝑠﷮𝑑𝑟﷯=−200 𝑟﷮−2﷯+4𝜋𝑟 Putting 𝑑𝑠﷮𝑑𝑟﷯=0 −200 𝑟﷮−2﷯+4𝜋𝑟=0 −200﷮ 𝑟﷮2﷯﷯+4𝜋𝑟=0 −200+4𝜋 𝑟﷮3﷯﷮ 𝑟﷮2﷯﷯=0 −200+4𝜋 𝑟﷮3﷯=0 4𝜋 𝑟﷮3﷯=200 𝜋 𝑟﷮3﷯= 200﷮4﷯ 𝜋𝑟﷮3﷯=50 𝑟﷮3﷯= 50﷮𝜋﷯﷯ 𝑟= 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯ Now, Finding 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯ 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯=−200 𝑟﷮−2﷯+4𝜋𝑟 Diff w.r.t 𝑟 𝑑𝑠﷮𝑑𝑟﷯= 𝑑 −200 𝑟﷮−2﷯+4𝜋𝑟﷯﷮𝑑𝑟﷯ 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯=−200 −2﷯ 𝑟﷮−2−1﷯+4𝜋 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯=400 𝑟﷮−3﷯+4𝜋 Putting value of 𝑟= 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯ 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯﷯﷮𝑟= 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯﷯= 400﷮ 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯﷯﷮3﷯﷯+4𝜋= 400﷮ 50﷮𝜋﷯﷯﷯+4𝜋 = 600𝜋﷮50﷯=12𝜋 > 0 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯>0 at 𝑟= 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯ Hence S is minimum at 𝑟= 50﷮𝜋﷯﷯﷮ 1﷮3﷯﷯ Now, Finding ℎ ℎ = 100﷮𝜋 𝑟﷮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 𝝅﷮𝟓𝟎﷯﷯﷮ 𝟐﷮𝟑﷯﷯ & 𝒉=𝟐 𝟓𝟎﷮𝝅﷯﷯﷮ 𝟏﷮𝟑﷯﷯ cm.