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

Transcript

Ex 6.5, 19 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. Let radius be r of the circle & let ๐‘ฅ be the length & ๐‘ฆ be the breadth of the rectangle Now, ฮ” ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 ๐‘ฅ^2+๐‘ฆ^2 = (2๐‘Ÿ)^2 ๐‘ฅ^2+๐‘ฆ^2= 4๐‘Ÿ2 ๐‘ฆ2 = 4๐‘Ÿ2 โ€“ ๐‘ฅ2 (As AC is diameter of circle) โ€ฆ(1) ๐‘ฆ= โˆš(4๐‘Ÿ"2 โ€“ " ๐‘ฅ"2" ) We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length ร— Breadth A = xy A = ๐‘ฅ โˆš(4๐‘Ÿ^2โˆ’๐‘ฅ^2 ) Since A has square root It will be difficult to differentiate So, we take Z = A2 Let Z = A2 Z = ๐‘ฅ^2ร— (โˆš(4๐‘Ÿ^2โˆ’๐‘ฅ^2 ))^2 Z = ๐‘ฅ^2ร—(4๐‘Ÿ^2โˆ’๐‘ฅ^2 ) Z = 4๐‘Ÿ^2 ๐‘ฅ^2โˆ’๐‘ฅ^4 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Diff. Z w.r.t ๐‘ฅ ๐‘‘Z/๐‘‘๐‘ฅ=๐‘‘(4๐‘Ÿ^2 ๐‘ฅ^2โˆ’๐‘ฅ^4 )/๐‘‘๐‘ฅ ๐‘‘Z/๐‘‘๐‘ฅ=4๐‘Ÿ^2ร—2๐‘ฅโˆ’4๐‘ฅ^3 ๐‘‘Z/๐‘‘๐‘ฅ=8๐‘Ÿ^2 ๐‘ฅโˆ’4๐‘ฅ^3 Putting ๐‘‘Z/๐‘‘๐‘ฅ = 0 8๐‘Ÿ^2 ๐‘ฅโˆ’4๐‘ฅ^3 = 0 4๐‘ฅ^3โˆ’8๐‘Ÿ^2 ๐‘ฅ = 0 ๐‘ฅ^3โˆ’2๐‘Ÿ^2 ๐‘ฅ = 0 ๐‘ฅ (๐‘ฅ^2โˆ’2๐‘Ÿ^2 ) = 0 Therefore, ๐‘ฅ = 0 Which is not possible ๐‘ฅ^2=2๐‘Ÿ^2 (๐‘‘^2 Z)/(๐‘‘๐‘ฅ^2 ) = 8๐‘Ÿ^2โˆ’12๐‘ฅ^2 Putting ๐’™^๐Ÿ=๐Ÿ๐’“^๐Ÿ (๐‘‘^2 Z)/(๐‘‘๐‘ฅ^2 ) = 8๐‘Ÿ^2โˆ’12ร—2 (๐‘‘^2 Z)/(๐‘‘๐‘ฅ^2 ) = 8๐‘Ÿ^2โˆ’24๐‘Ÿ^2 (๐‘‘^2 Z)/(๐‘‘๐‘ฅ^2 ) = โˆ’16๐‘Ÿ^2 < 0 Hence, (๐‘‘^2 Z)/(๐‘‘๐‘ฅ^2 ) < 0 at ๐‘ฅ^2=2๐‘Ÿ^2 Thus area is maximum when ๐‘ฅ^2=2๐‘Ÿ^2 Now, finding y ๐‘ฆ = โˆš(4๐‘Ÿ^2 โˆ’๐‘ฅ^2 ) Putting ๐‘ฅ^2=2๐‘Ÿ^2 ๐‘ฆ = โˆš(4๐‘Ÿ^2 โˆ’ใ€–2๐‘Ÿใ€—^2 ) ๐‘ฆ = โˆš(ใ€–2๐‘Ÿใ€—^2 ) ๐‘ฆ = โˆš2 ๐‘Ÿ Therefore ๐‘ฅ = ๐‘ฆ = โˆš2 ๐‘Ÿ Hence area is maximum when ๐’™ = ๐’š โˆด The rectangle is a square.

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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.