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Misc 11 - A window is a rectangle surmounted by semicircular - Minima/ maxima (statement questions) - Geometry questions

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  1. Class 12
  2. Important Question for exams Class 12
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Misc 11 A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. Let Length of rectangle be x & breadth of rectangle be y Diameter of semicircle = x ∴ Radius of semicircle = 𝑥﷮2﷯ Given , Perimeter of window = 10 m Length + 2 × Breadth + circumference of semicircle = 10 𝑥+2𝑦+𝜋 𝑥﷮2﷯﷯=10 2𝑦=10−𝑥− 𝜋𝑥﷮2﷯ 𝑦=5− 𝑥﷮2﷯ − 𝜋𝑥﷮2﷯ 𝑦=5−𝑥 1﷮2﷯+ 𝜋﷮4﷯﷯ We need to maximize area of window Area of window = Area of rectangle + Area of Semicircle A = 𝑥𝑦+ 1﷮2﷯×𝜋 𝜋﷮2﷯﷯﷮2﷯ A = 𝑥 5−𝑥 1﷮2﷯+ 𝜋﷮4﷯﷯﷯+ 𝜋 𝑥﷮2﷯﷮8﷯ A = 5𝑥− 5﷮2﷯ 𝑥﷮2﷯− 𝜋 𝑥﷮2﷯﷮4﷯+ 𝜋 𝑥﷮2﷯﷮8﷯ A = 5𝑥− 5﷮2﷯ 𝑥﷮2﷯− 𝜋 𝑥﷮2﷯﷮8﷯ Finding 𝑑𝐴﷮𝑑𝑥﷯ 𝑑𝐴﷮𝑑𝑥﷯= 𝑑 5𝑥 − 5﷮2﷯ 𝑥﷮2﷯ − 𝜋 𝑥﷮2﷯﷮8﷯﷯﷮𝑑𝑥﷯ 𝑑𝐴﷮𝑑𝑥﷯=5−5𝑥− 𝜋𝑥﷮4﷯ Putting 𝑑𝐴﷮𝑑𝑥﷯=0 0 = 5−5𝑥− 𝜋𝑥﷮4﷯ 5− 𝜋﷮4﷯﷯𝑥=5 𝑥= 5﷮ 5 − 𝜋﷮4﷯﷯﷯ 𝑥= 20﷮𝜋 + 4﷯ Now, calculating 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯ at 𝑥= 20﷮𝜋 + 4﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯= 𝑑 5 − 5𝑥 − 𝜋𝑥﷮4﷯﷯﷮𝑑𝑥﷯ 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯=− 5− 𝜋﷮4﷯ <0 So, 𝑑﷮2﷯𝐴﷮𝑑 𝑥﷮2﷯﷯<0 at 𝑥= 20﷮𝜋 + 4﷯ Hence, 𝑥= 20﷮𝜋 + 4﷯ maxima Hence, A is maximum when 𝑥= 20﷮𝜋 + 4﷯ Now, Finding value of y 𝑦=5−𝑥 1﷮2﷯+ 𝜋﷮4﷯﷯ 𝑦=5− 20﷮𝜋 + 4﷯ 1﷮2﷯+ 𝜋﷮4﷯﷯ 𝑦=5− 20﷮𝜋 + 4﷯ 4 + 2𝜋﷮2﷯﷯ 𝑦=5− 5﷮2﷯ 4 + 2𝜋﷯﷮𝜋 + 4﷯ 𝑦=5−5 2 + 𝜋﷯﷮𝜋 + 4﷯ 𝑦=5 𝜋 + 4 − 2 + 𝜋﷯﷮𝜋 + 4﷯﷯ 𝑦= 5 × 2﷮𝜋 + 4﷯ 𝑦= 10﷮𝜋 + 4﷯ Hence, for maximum area, Length = 𝒙= 𝟐𝟎﷮𝝅 + 𝟒﷯ m & Breadth = 𝒚= 𝟏𝟎﷮𝝅 + 𝟒﷯ m

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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