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Last updated at Jan. 7, 2020 by Teachoo

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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 ๐ฅ be the length & ๐ฆ be the breadth of the rectangle Now, ฮ ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 ๐ฅ^2+๐ฆ^2 = (2๐)^2 ๐ฅ^2+๐ฆ^2= 4๐2 ๐ฆ2 = 4๐2 โ ๐ฅ2 (As AC is diameter of circle) โฆ(1) ๐ฆ= โ(4๐"2 โ " ๐ฅ"2" ) We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length ร Breadth A = xy A = ๐ฅ โ(4๐^2โ๐ฅ^2 ) Since A has square root It will be difficult to differentiate So, we take Z = A2 Let Z = A2 Z = ๐ฅ^2ร (โ(4๐^2โ๐ฅ^2 ))^2 Z = ๐ฅ^2ร(4๐^2โ๐ฅ^2 ) Z = 4๐^2 ๐ฅ^2โ๐ฅ^4 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Diff. Z w.r.t ๐ฅ ๐Z/๐๐ฅ=๐(4๐^2 ๐ฅ^2โ๐ฅ^4 )/๐๐ฅ ๐Z/๐๐ฅ=4๐^2ร2๐ฅโ4๐ฅ^3 ๐Z/๐๐ฅ=8๐^2 ๐ฅโ4๐ฅ^3 Putting ๐Z/๐๐ฅ = 0 8๐^2 ๐ฅโ4๐ฅ^3 = 0 4๐ฅ^3โ8๐^2 ๐ฅ = 0 ๐ฅ^3โ2๐^2 ๐ฅ = 0 ๐ฅ (๐ฅ^2โ2๐^2 ) = 0 Therefore, ๐ฅ = 0 Which is not possible ๐ฅ^2=2๐^2 (๐^2 Z)/(๐๐ฅ^2 ) = 8๐^2โ12๐ฅ^2 Putting ๐^๐=๐๐^๐ (๐^2 Z)/(๐๐ฅ^2 ) = 8๐^2โ12ร2 (๐^2 Z)/(๐๐ฅ^2 ) = 8๐^2โ24๐^2 (๐^2 Z)/(๐๐ฅ^2 ) = โ16๐^2 < 0 Hence, (๐^2 Z)/(๐๐ฅ^2 ) < 0 at ๐ฅ^2=2๐^2 Thus area is maximum when ๐ฅ^2=2๐^2 Now, finding y ๐ฆ = โ(4๐^2 โ๐ฅ^2 ) Putting ๐ฅ^2=2๐^2 ๐ฆ = โ(4๐^2 โใ2๐ใ^2 ) ๐ฆ = โ(ใ2๐ใ^2 ) ๐ฆ = โ2 ๐ Therefore ๐ฅ = ๐ฆ = โ2 ๐ Hence area is maximum when ๐ = ๐ โด The rectangle is a square.

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.