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


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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Ex 6.5, 22 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? A wire of length 28m cut into two parts Let ๐‘ฅ be the length of first part Length of 2nd part = (28โˆ’๐‘ฅ)๐‘š Given that one part is converted into a square Let length of ๐‘ฅ m be converted into a square Thus, Perimeter of square =๐‘ฅ 4(๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘“ ๐‘ ๐‘ž๐‘ข๐‘Ž๐‘Ÿ๐‘’ )=๐‘ฅ side of square of =๐‘ฅ/4 Also, other part is converted into Circle โ‡’ (28โˆ’๐‘ฅ) m length converted into a Circle of radius ๐‘Ÿ Circumference of circle = (28โˆ’๐‘ฅ) m 2๐œ‹๐‘Ÿ=(28โˆ’๐‘ฅ) ๐‘Ÿ=((28 โˆ’ ๐‘ฅ)/2๐œ‹) m We need to find length of pieces so that the combined Area of Circle & Square is minimum Let T be the total area of circle & square T = Area of Circle + Area of Square T = ๐œ‹๐‘Ÿ^2+(๐‘ ๐‘–๐‘‘๐‘’)^2 T = ๐œ‹((28 โˆ’ ๐‘ฅ)/2๐œ‹)^2+๐‘ฅ^2/16 We have to minimize T Diff T w.r.t ๐‘ฅ ๐‘‘๐‘‡/๐‘‘๐‘ฅ=๐‘‘/๐‘‘๐‘ฅ ((28 โˆ’ ๐‘ฅ)^2/4๐œ‹+๐‘ฅ^2/16 " " ) = 2(28 โˆ’ ๐‘ฅ)/4๐œ‹ . ๐‘‘(28 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ+2๐‘ฅ/16 = (28 โˆ’ ๐‘ฅ)/2๐œ‹ (โˆ’1)+๐‘ฅ/8 Putting ๐’…๐‘ป/๐’…๐’™=๐ŸŽ (๐‘ฅ โˆ’ 28)/2๐œ‹+๐‘ฅ/8=0 (4(๐‘ฅ โˆ’ 28) + ๐œ‹๐‘ฅ)/8๐œ‹ =0 4x โ€“ 112 +๐œ‹ ๐‘ฅ=0ร—8 ๐œ‹ ๐‘ฅ(4+๐œ‹)=112 ๐‘ฅ=112/(4 + ๐œ‹) Now finding (๐’…^๐Ÿ ๐‘ป)/(๐’…๐’™^๐Ÿ ) ๐‘‘๐‘‡/๐‘‘๐‘ฅ=(โˆ’28 + ๐‘ฅ)/2๐œ‹+๐‘ฅ/8 (๐‘‘^2 ๐‘‡)/(๐‘‘๐‘ฅ^2 )=1/2๐œ‹ (0+1)+1/8 (๐‘‘^2 ๐‘‡)/(๐‘‘๐‘ฅ^2 )=1/2๐œ‹+1/8 > 0 Hence (๐‘‘^2 ๐‘‡)/(๐‘‘๐‘ฅ^2 )>0 at ๐‘ฅ=112/(4 + ๐œ‹) โˆด Total area is Minimum at ๐‘ฅ = 112/(4 + ๐œ‹) Finding length of other part Length of other part = 28 โ€“ x = 28 โ€“ 112/(4 + ๐œ‹) = (28(4 + ๐œ‹)โˆ’112)/(4 + ๐œ‹)= (28 ๐œ‹)/(4 + ๐œ‹) โˆด Length of first part is ๐Ÿ๐Ÿ๐Ÿ/(๐Ÿ’ + ๐…) & another part is ๐Ÿ๐Ÿ–๐…/(๐Ÿ’ + ๐…)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.