Ex 6.5, 19 - Show that of all rectangles inscribed in circle - Minima/ maxima (statement questions) - Geometry questions

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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 𝑥 be the length & 𝑦 be the breadth of the rectangle Now, Δ ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 𝑥﷮2﷯+ 𝑦﷮2﷯ = 2𝑟﷯﷮2﷯ 𝑥﷮2﷯+ 𝑦﷮2﷯= 4𝑟2 𝑦2 = 4𝑟2 – 𝑥2 𝑦= ﷮4𝑟2 – 𝑥2﷯ We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length × Breadth A = xy A = 𝑥 ﷮4𝑟2 – 𝑥2﷯ Diff. w.r.t 𝑥 𝑑𝐴﷮𝑑𝑥﷯= 𝑑 4𝑥 ﷮4𝑟2 – 𝑥2﷯﷯﷮𝑑𝑥﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . ﷮4𝑟2 – 𝑥2﷯ + 𝑑 ﷮4𝑟2 – 𝑥2﷯﷯﷮𝑑𝑥﷯ . 𝑥﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 1 . ﷮4𝑟2 – 𝑥2﷯+ 1﷮2 ﷮4𝑟2 – 𝑥2﷯﷯ . 𝑑﷮𝑑𝑥﷯ 4𝑟2 – 𝑥2﷯ . 𝑥﷯ 𝑑𝐴﷮𝑑𝑥﷯ = ﷮4𝑟2 – 𝑥2 ﷯− 2 𝑥﷮2﷯﷮2 ﷮4 𝑟﷮2﷯− 𝑥﷮2﷯﷯﷯ ﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 2 ﷮4𝑟2 – 𝑥2﷯﷯﷮2﷯− 2 𝑥﷮2﷯﷮2 ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 2 4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ − 2 𝑥﷮2﷯﷮2 ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 4 𝑟﷮2﷯− 𝑥﷮2﷯ − 𝑥﷮2﷯﷮ ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 4 𝑟﷮2﷯ − 2𝑥﷮2﷯﷮ ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 2 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷮ ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ Putting 𝑑𝐴﷮𝑑𝑥﷯ = 0 4 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷮ ﷮4𝑟2 – 𝑥2﷯﷯ ﷯=0 2 𝑟﷮2﷯ − 𝑥﷮2﷯ = 0 × ﷮4𝑟2 – 𝑥2﷯﷮4﷯ 2 𝑟﷮2﷯ − 𝑥﷮2﷯ = 0 − 𝑥﷮2﷯ = –2 𝑟﷮2﷯ 𝑥﷮2﷯ = 2 𝑟﷮2﷯ 𝑥 = ﷮2 𝑟﷮2﷯﷯ 𝑥 = ﷮2﷯ 𝑟 Finding 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ 𝑑𝐴﷮𝑑𝑥﷯ = 2 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷮ ﷮4𝑟2 – 𝑥2﷯﷯ ﷯ Diff w.r.t 𝑥 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 2 . 𝑑﷮𝑑𝑥﷯ 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷮ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯ ﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 2 𝑑 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮𝑑𝑥﷯ . ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯− 𝑑﷮𝑑𝑥﷯ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯ . 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ ﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 0 −2𝑥﷯ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯ − 1﷮2 ﷮4𝑟2 – 𝑥2﷯﷯ . 𝑑﷮𝑑𝑥﷯ 4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ . 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ ﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 −2𝑥 ﷮4𝑟2 – 𝑥2﷯ − 1﷮2 ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯ . 0 −2𝑥﷯ . 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ ﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 −2𝑥 ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯+ 𝑥 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯﷮ 4𝑟2 – 𝑥2﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 −2𝑥 ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ ﷯﷮2﷯+ 𝑥 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ 4𝑟2 – 𝑥2﷯﷮2﷯ ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 −2𝑥 4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ + 𝑥 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ 4𝑟2 – 𝑥2﷯﷮1+ 1﷮2﷯﷯﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 8 −2𝑥 4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯+ 𝑥 2 𝑟﷮2﷯ − 𝑥﷮2﷯﷯﷮ 4𝑟2 – 𝑥2﷯﷮ 3﷮2﷯﷯﷯﷯ Putting 𝑥 = ﷮2﷯ 𝑟 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯│﷮ 𝑥﷮2﷯= 2 𝑟﷮2﷯﷯ = 8 −2𝑥 ﷮2𝑟﷯﷯ 4 𝑟﷮2﷯ − 2𝑟﷮2﷯﷯+ ﷮2𝑟﷯ 2 𝑟﷮2﷯ − 2𝑟﷮2﷯﷯﷮ 4 𝑟﷮2﷯ − 2𝑟﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷯ = 8 −2 ﷮2𝑟 ﷯ 2 r﷮2﷯﷯+ ﷮2𝑟﷯ 0﷯﷮ 2 𝑟﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷯ = 8 −2 ﷮2𝑟 ﷯ 2 r﷮2﷯﷯ +0﷮ 2 r﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷯ = 8 −4 ﷮2𝑟﷯ 𝑟﷮2﷯﷮2 ﷮2﷯ 𝑟﷮3﷯﷯﷯ = 16 ﷮𝑟﷯﷮𝑟﷯ = −16﷮𝑟﷯ < 0 Hence 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ < 0 at 𝑥 = ﷮2𝑟﷯ Thus area is maximum when 𝑥 = ﷮2﷯ 𝑟 Now, finding y 𝑦 = ﷮4 𝑟﷮2﷯ − 𝑥﷮2﷯﷯ Putting 𝑥﷮2﷯ = ﷮2𝑟﷯ 𝑦 = ﷮4 𝑟﷮2﷯ − ﷮2𝑟﷯ ﷯﷮2﷯﷯ 𝑦 = ﷮4 𝑟﷮2﷯ − 2𝑟﷮2﷯﷯ 𝑦 = ﷮ 2𝑟﷮2﷯﷯ 𝑦 = ﷮2﷯ 𝑟 Therefore 𝑥 = 𝑦 = ﷮2﷯ 𝑟 Hence area is maximum when 𝒙 = 𝒚 ∴ The rectangle is a square.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.