# Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

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 ๐ฅ be the length of first part Length of 2nd part = (28โ๐ฅ)๐ Given that one part is converted into a square Let length of ๐ฅ m be converted into a square Thus, Perimeter of square =๐ฅ 4(๐ ๐๐๐ ๐๐ ๐ ๐๐ข๐๐๐ )=๐ฅ side of square of =๐ฅ/4 Also, other part is converted into Circle โ (28โ๐ฅ) m length converted into a Circle of radius ๐ Circumference of circle = (28โ๐ฅ) m 2๐๐=(28โ๐ฅ) ๐=((28 โ ๐ฅ)/2๐) 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 = ๐๐^2+(๐ ๐๐๐)^2 T = ๐((28 โ ๐ฅ)/2๐)^2+๐ฅ^2/16 We have to minimize T Diff T w.r.t ๐ฅ ๐๐/๐๐ฅ=๐/๐๐ฅ ((28 โ ๐ฅ)^2/4๐+๐ฅ^2/16 " " ) = 2(28 โ ๐ฅ)/4๐ . ๐(28 โ ๐ฅ)/๐๐ฅ+2๐ฅ/16 = (28 โ ๐ฅ)/2๐ (โ1)+๐ฅ/8 Putting ๐ ๐ป/๐ ๐=๐ (๐ฅ โ 28)/2๐+๐ฅ/8=0 (4(๐ฅ โ 28) + ๐๐ฅ)/8๐ =0 4x โ 112 +๐ ๐ฅ=0ร8 ๐ ๐ฅ(4+๐)=112 ๐ฅ=112/(4 + ๐) Now finding (๐ ^๐ ๐ป)/(๐ ๐^๐ ) ๐๐/๐๐ฅ=(โ28 + ๐ฅ)/2๐+๐ฅ/8 (๐^2 ๐)/(๐๐ฅ^2 )=1/2๐ (0+1)+1/8 (๐^2 ๐)/(๐๐ฅ^2 )=1/2๐+1/8 > 0 Hence (๐^2 ๐)/(๐๐ฅ^2 )>0 at ๐ฅ=112/(4 + ๐) โด Total area is Minimum at ๐ฅ = 112/(4 + ๐) Finding length of other part Length of other part = 28 โ x = 28 โ 112/(4 + ๐) = (28(4 + ๐)โ112)/(4 + ๐)= (28 ๐)/(4 + ๐) โด Length of first part is ๐๐๐/(๐ + ๐ ) & another part is ๐๐๐ /(๐ + ๐ )

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.