# Misc 10 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 10 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle. Let 𝑥 be the radius of circle & y be the side of the square Given sum of perimeter of circle & square is k i.e. Circumference of circle + Perimeter of square = k i.e. 2π 𝑟𝑎𝑑𝑖𝑢𝑠+ 4 𝑠𝑖𝑑𝑒 = k 2π𝑥 + 4y = k 4𝑦 = k – 2π 𝑥 y = 𝑘 − 2𝜋𝑥4 We have to prove that sum of their area is least when side of square = 2 𝑟𝑎𝑑𝑖𝑢𝑠 i.e. we need to minimize the sum of areas Let A be then sum of their area A = Area of circle + Area of square A = π 𝑟𝑎𝑑𝑖𝑢𝑠2+ 𝑠𝑖𝑑𝑒2 A = π𝑥2 + 𝑦2 A = π𝑥2 + 𝑘 − 2𝜋𝑥42 Differentiating w.r.t 𝑥 𝑑𝐴𝑑𝑥= 𝑑 𝜋𝑥2 + 𝑘 − 2𝜋𝑥42𝑑𝑥 = 𝑑 𝜋 𝑥2𝑑𝑥+ 𝑑𝑑𝑥 𝑘 − 2𝜋𝑥42 = π 𝑑 𝑥2𝑑𝑥+ 1 42 𝑑 𝑘 − 2𝜋𝑥2𝑑𝑥 = π 2𝑥+ 1162 𝑘−2𝜋𝑥. 𝑑 𝑘 − 2𝜋𝑥𝑑𝑥 = 2π 𝑥 + 1162 𝑘−2𝜋𝑥. 0−2𝜋 = 2π 𝑥 + 18 𝑘−2𝜋𝑥 −2𝜋 = 2π 𝑥 – 𝜋 𝑘 − 2𝜋𝑥4 =2𝜋𝑥− 𝜋4 𝑘−2𝜋𝑥 = 8𝜋𝑥 − 𝜋𝑘 + 2 𝜋2𝑥4 = 2𝜋𝑥 4 + 𝜋 − 𝜋𝑘4 Putting 𝑑𝐴𝑑𝑥=0 2𝜋𝑥 4 + 𝜋 − 𝜋𝑘4=0 2π𝑥 4+𝜋=𝜋𝑘 𝑥 = 𝜋𝑘2𝜋 4 + 𝜋 𝑥 = 𝑘2 4 + 𝜋 Finding 𝑑2𝐴 𝑑′ 𝑥2 𝑑𝐴𝑑𝑥= 2𝜋𝑥 4 + 𝜋 − 𝜋𝑘4 𝑑𝐴𝑑𝑥= 2𝜋𝑥 4 + 𝜋4 − 𝜋𝑘 4 Differentiating w.r.t 𝑥 𝑑2𝐴𝑑 𝑥2 = 𝑑𝑑𝑥 2𝜋 4 + 𝜋4.𝑥− 𝑑𝑑𝑥 𝜋𝑘4 𝑑2𝐴𝑑 𝑥2 = 2π 4 + 𝜋4 . 𝑑 𝑥𝑑𝑥−0 𝑑2𝐴𝑑 𝑥2 = 2𝜋 4 + 𝜋4 > 0 ⇒ 𝑑2𝐴𝑑 𝑥2 >0 at 𝑥 = 𝑘2 4 + 𝜋 ∴ 𝑥 = 𝑘2 4 + 𝜋 is point of minima Thus, area is least when 𝑥 = 𝑘2 4 + 𝜋 From (1) 𝑦 = 𝑘 − 2𝜋𝑥4 Putting value of 𝑥 = 𝑘2 4 + 𝜋 𝑦 = 𝑘 − 2𝜋 𝑘2 4 + 𝜋4 𝑦 = 𝑘4 1− 2𝜋2 4 + 𝜋 𝑦 = 𝑘4 1− 𝜋4 + 𝜋 𝑦 = 𝑘4 4 + 𝜋 + 𝜋4 + 𝜋 𝑦 = 𝑘4 44 + 𝜋 𝑦 = 𝑘4 + 𝜋 Thus, 𝑦=2 𝑘2 𝜋 + 4 𝑦=2𝑥 Hence, proved sum of their area is least when 𝒚 = 𝟐𝒙

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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Davneet Singh

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.