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
Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 6
Ex 6.5,19 - Chapter 6 Class 12 Application of Derivatives - Part 7

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  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 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
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.