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Ex 6.5, 19 - Show that of all rectangles inscribed in a fixed circle,

Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 2
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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 π‘₯ be the length & 𝑦 be the breadth of the rectangle Now, Ξ” ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 π‘₯^2+𝑦^2 = (2π‘Ÿ)^2 π‘₯^2+𝑦^2= 4π‘Ÿ2 𝑦2 = 4π‘Ÿ2 – π‘₯2 (As AC is diameter of circle) 𝑦= √(4π‘Ÿ"2 – " π‘₯"2" ) We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length Γ— Breadth A = xy A = π‘₯ √(4π‘Ÿ^2βˆ’π‘₯^2 ) Since A has square root It will be difficult to differentiate So, we take Z = A2 Let Z = A2 Z = π‘₯^2Γ— (√(4π‘Ÿ^2βˆ’π‘₯^2 ))^2 Z = π‘₯^2Γ—(4π‘Ÿ^2βˆ’π‘₯^2 ) Z = 4π‘Ÿ^2 π‘₯^2βˆ’π‘₯^4 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Diff. Z w.r.t π‘₯ 𝑑Z/𝑑π‘₯=𝑑(4π‘Ÿ^2 π‘₯^2βˆ’π‘₯^4 )/𝑑π‘₯ 𝑑Z/𝑑π‘₯=4π‘Ÿ^2Γ—2π‘₯βˆ’4π‘₯^3 𝑑Z/𝑑π‘₯=8π‘Ÿ^2 π‘₯βˆ’4π‘₯^3 Putting 𝑑Z/𝑑π‘₯ = 0 8π‘Ÿ^2 π‘₯βˆ’4π‘₯^3 = 0 4π‘₯^3βˆ’8π‘Ÿ^2 π‘₯ = 0 π‘₯^3βˆ’2π‘Ÿ^2 π‘₯ = 0 π‘₯ (π‘₯^2βˆ’2π‘Ÿ^2 ) = 0 Therefore, π‘₯ = 0 Which is not possible Finding (𝒅^𝟐 𝐙)/(𝐝𝒙^𝟐 ) 𝑑Z/𝑑π‘₯=8π‘Ÿ^2 π‘₯βˆ’4π‘₯^3 Diff w.r.t π‘₯ (𝑑^2 Z)/(𝑑π‘₯^2 ) = 𝑑/𝑑π‘₯ [8π‘Ÿ^2 π‘₯βˆ’4π‘₯^3 ] (𝑑^2 Z)/(𝑑π‘₯^2 ) = 8π‘Ÿ^2βˆ’4Γ—3π‘₯^2 (𝑑^2 Z)/(𝑑π‘₯^2 ) = 8π‘Ÿ^2βˆ’12π‘₯^2 Putting 𝒙^𝟐=πŸπ’“^𝟐 (𝑑^2 Z)/(𝑑π‘₯^2 ) = 8π‘Ÿ^2βˆ’12Γ—2π‘Ÿ^2 (𝑑^2 Z)/(𝑑π‘₯^2 ) = 8π‘Ÿ^2βˆ’24π‘Ÿ^2 (𝑑^2 Z)/(𝑑π‘₯^2 ) = βˆ’16π‘Ÿ^2 < 0 Hence, (𝑑^2 Z)/(𝑑π‘₯^2 ) < 0 at π‘₯^2=2π‘Ÿ^2 Thus area is maximum when π‘₯^2=2π‘Ÿ^2 Now, finding y 𝑦 = √(4π‘Ÿ^2 βˆ’π‘₯^2 ) Putting π‘₯^2=2π‘Ÿ^2 𝑦 = √(4π‘Ÿ^2 βˆ’γ€–2π‘Ÿγ€—^2 ) 𝑦 = √(γ€–2π‘Ÿγ€—^2 ) 𝑦 = √2 π‘Ÿ Therefore π‘₯ = 𝑦 = √2 π‘Ÿ Hence area is maximum when 𝒙 = π’š ∴ The rectangle is a square.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.