Misc 10 - Sum of perimeter of a circle and square is k - Minima/ maxima (statement questions) - Geometry questions

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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 − 2𝜋𝑥﷮4﷯ 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 𝑟𝑎𝑑𝑖𝑢𝑠﷯﷮2﷯+ 𝑠𝑖𝑑𝑒﷯﷮2﷯ A = π𝑥2 + 𝑦2 A = π𝑥2 + 𝑘 − 2𝜋𝑥﷮4﷯﷯﷮2﷯ Differentiating w.r.t 𝑥 𝑑𝐴﷮𝑑𝑥﷯= 𝑑 𝜋𝑥﷮2﷯ + 𝑘 − 2𝜋𝑥﷮4﷯﷯﷮2﷯﷯﷮𝑑𝑥﷯ = 𝑑 𝜋 𝑥﷮2﷯﷯﷮𝑑𝑥﷯+ 𝑑﷮𝑑𝑥﷯ 𝑘 − 2𝜋𝑥﷮4﷯﷯﷮2﷯ = π 𝑑 𝑥﷮2﷯﷯﷮𝑑𝑥﷯+ 1﷮ 4﷮2﷯﷯ 𝑑 𝑘 − 2𝜋𝑥﷯﷮2﷯﷮𝑑𝑥﷯ = π 2𝑥﷯+ 1﷮16﷯2 𝑘−2𝜋𝑥﷯. 𝑑 𝑘 − 2𝜋𝑥﷯﷮𝑑𝑥﷯ = 2π 𝑥 + 1﷮16﷯2 𝑘−2𝜋𝑥﷯. 0−2𝜋﷯ = 2π 𝑥 + 1﷮8﷯ 𝑘−2𝜋𝑥﷯ −2𝜋﷯ = 2π 𝑥 – 𝜋 𝑘 − 2𝜋𝑥﷯﷮4﷯ =2𝜋𝑥− 𝜋﷮4﷯ 𝑘−2𝜋𝑥﷯ = 8𝜋𝑥 − 𝜋𝑘 + 2 𝜋﷮2﷯𝑥﷮4﷯ = 2𝜋𝑥 4 + 𝜋﷯ − 𝜋𝑘﷮4﷯ Putting 𝑑𝐴﷮𝑑𝑥﷯=0 2𝜋𝑥 4 + 𝜋﷯ − 𝜋𝑘﷮4﷯=0 2π𝑥 4+𝜋﷯=𝜋𝑘 𝑥 = 𝜋𝑘﷮2𝜋 4 + 𝜋﷯﷯ 𝑥 = 𝑘﷮2 4 + 𝜋﷯﷯ Finding 𝑑﷮2﷯𝐴﷮ 𝑑﷮′﷯ 𝑥﷮2﷯﷯ 𝑑𝐴﷮𝑑𝑥﷯= 2𝜋𝑥 4 + 𝜋﷯ − 𝜋𝑘﷮4﷯ 𝑑𝐴﷮𝑑𝑥﷯= 2𝜋𝑥 4 + 𝜋﷯﷮4﷯ − 𝜋𝑘 ﷮4﷯ Differentiating w.r.t 𝑥 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 𝑑﷮𝑑𝑥﷯ 2𝜋 4 + 𝜋﷯﷮4﷯.𝑥﷯− 𝑑﷮𝑑𝑥﷯ 𝜋𝑘﷮4﷯﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 2π 4 + 𝜋﷯﷮4﷯ . 𝑑 𝑥﷯﷮𝑑𝑥﷯−0 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ = 2𝜋 4 + 𝜋﷯﷮4﷯ > 0 ⇒ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ >0 at 𝑥 = 𝑘﷮2 4 + 𝜋﷯﷯ ∴ 𝑥 = 𝑘﷮2 4 + 𝜋﷯﷯ is point of minima Thus, area is least when 𝑥 = 𝑘﷮2 4 + 𝜋﷯﷯ From (1) 𝑦 = 𝑘 − 2𝜋𝑥﷮4﷯ Putting value of 𝑥 = 𝑘﷮2 4 + 𝜋﷯﷯ 𝑦 = 𝑘 − 2𝜋 𝑘﷮2 4 + 𝜋﷯﷯﷯﷮4﷯ 𝑦 = 𝑘﷮4﷯ 1− 2𝜋﷮2 4 + 𝜋﷯﷯﷯ 𝑦 = 𝑘﷮4﷯ 1− 𝜋﷮4 + 𝜋﷯﷯ 𝑦 = 𝑘﷮4﷯ 4 + 𝜋 + 𝜋﷮4 + 𝜋﷯﷯ 𝑦 = 𝑘﷮4﷯ 4﷮4 + 𝜋﷯﷯ 𝑦 = 𝑘﷮4 + 𝜋﷯ Thus, 𝑦=2 𝑘﷮2 𝜋 + 4﷯﷯﷯ 𝑦=2𝑥 Hence, proved sum of their area is least when 𝒚 = 𝟐𝒙

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