Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12







Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
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 = ๐๐^2 โ V = ๐๐^2 โ 100 = ๐๐^2 โ 100/(๐๐^2 )=โ โ=(100/(๐๐^2 )) We need to minimize surface area We know that Surface area of Cylinder = 2ฯ๐h + 2ฯ๐2 S = 2๐๐โ + 2๐๐2 Putting value of h =(100/(๐๐^2 โ)) S = 2ฯr (100/(๐๐^2 ))+"2ฯr2 " S = 2(100/๐)+"2ฯr2 " S = 200/๐+2๐๐^2 S = 200r โ1 + 2๐๐^2 We need to Minimize S S = 200r โ1 + 2๐๐^2 Diff w.r.t.๐ ๐๐ /๐๐=๐(200"r โ1 + " 2๐๐^2 )/๐๐ ๐๐ /๐๐=200(โ1) ๐^(โ1โ1)+2๐ ร2๐^(2โ1) ๐๐ /๐๐=โ200๐^(โ2)+4๐๐ Putting ๐ ๐/๐ ๐=๐ โ200๐^(โ2)+4๐๐=0 (โ200)/๐^2 +4๐๐=0 (โ200+4๐๐^3)/๐^2 =0 โ200+4๐๐^3=0 4๐๐^3=200 ๐๐^3=200/4 ใ๐๐ใ^3=50 ๐^3=(50/๐) ๐=(50/๐)^(1/3) Now, Finding (๐ ^๐ ๐)/(๐ ๐^๐ ) (๐^2 ๐ )/(๐๐^2 )=โ200๐^(โ2)+4๐๐ Diff w.r.t ๐ ๐๐ /๐๐=๐(โ200๐^(โ2)+4๐๐)/๐๐ (๐^2 ๐ )/(๐๐^2 )=โ200(โ2) ๐^(โ2โ1)+4๐ (๐^2 ๐ )/(๐๐^2 )=400๐^(โ3)+4๐ Putting value of ๐=(50/๐)^(1/3) โ (๐^2 ๐ )/(๐๐^2 )โค|_(๐=(50/๐)^(1/3) )=400/((50/๐)^(1/3) )^3 +4๐=400/((50/๐) )+4๐ =600๐/50=12๐ > 0 (๐^2 ๐ )/(๐๐^2 )>0 at ๐=(50/๐)^(1/3) Hence S is minimum at ๐=(50/๐)^(1/3) Now, Finding โ โ = (100/(๐๐^2 )) Putting value of ๐=(50/๐)^(1/3) cm โ = 100/ใ๐[(50/๐)^(1/3) ]ใ^2 = 100/ใ๐(50/๐)ใ^(2/3) = 100/๐ (๐/50)^(2/3) = 100/๐ ร ๐^(2/3)/ใ50ใ^(2/3) = (2 ร 50)/ใ50ใ^(2/3) ร ๐^(2/3)/๐ = (2 ร ใ50ใ^(1 โ 2/3))/๐^(1 โ 2/3) = 2 ร ใ50ใ^(1/3)/๐^(1/3) = 2(50/๐)^(1/3) Hence total surface area is least when Radius of base is (๐๐/๐)^(๐/๐) & ๐=๐ (๐๐/๐ )^(๐/๐) cm.
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