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Minima/ maxima (statement questions) - Geometry questions
Ex 6.5, 27 (MCQ)
Example 35 Important
Example 41 Important
Example 36
Misc 12 Important
Example 37 Important
Misc 9 Important
Ex 6.5,21 You are here
Ex 6.5, 20 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5,22 Important
Misc 10
Misc 11 Important
Ex 6.5,17
Ex 6.5,18 Important
Example 50 Important
Ex 6.5,19 Important
Misc 8 Important
Ex 6.5,23 Important
Misc 15 Important
Misc 17 Important
Example 38 Important
Misc 18 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at April 15, 2021 by Teachoo
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 = 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.