# Misc 11

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

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 = 𝑥2 Given , Perimeter of window = 10 m Length + 2 × Breadth + circumference of semicircle = 10 𝑥+2𝑦+𝜋 𝑥2=10 2𝑦=10−𝑥− 𝜋𝑥2 𝑦=5− 𝑥2 − 𝜋𝑥2 𝑦=5−𝑥 12+ 𝜋4 We need to maximize area of window Area of window = Area of rectangle + Area of Semicircle A = 𝑥𝑦+ 12×𝜋 𝜋22 A = 𝑥 5−𝑥 12+ 𝜋4+ 𝜋 𝑥28 A = 5𝑥− 52 𝑥2− 𝜋 𝑥24+ 𝜋 𝑥28 A = 5𝑥− 52 𝑥2− 𝜋 𝑥28 Finding 𝑑𝐴𝑑𝑥 𝑑𝐴𝑑𝑥= 𝑑 5𝑥 − 52 𝑥2 − 𝜋 𝑥28𝑑𝑥 𝑑𝐴𝑑𝑥=5−5𝑥− 𝜋𝑥4 Putting 𝑑𝐴𝑑𝑥=0 0 = 5−5𝑥− 𝜋𝑥4 5− 𝜋4𝑥=5 𝑥= 5 5 − 𝜋4 𝑥= 20𝜋 + 4 Now, calculating 𝑑2𝐴𝑑 𝑥2 at 𝑥= 20𝜋 + 4 𝑑2𝐴𝑑 𝑥2= 𝑑 5 − 5𝑥 − 𝜋𝑥4𝑑𝑥 𝑑2𝐴𝑑 𝑥2=− 5− 𝜋4 <0 So, 𝑑2𝐴𝑑 𝑥2<0 at 𝑥= 20𝜋 + 4 Hence, 𝑥= 20𝜋 + 4 maxima Hence, A is maximum when 𝑥= 20𝜋 + 4 Now, Finding value of y 𝑦=5−𝑥 12+ 𝜋4 𝑦=5− 20𝜋 + 4 12+ 𝜋4 𝑦=5− 20𝜋 + 4 4 + 2𝜋2 𝑦=5− 52 4 + 2𝜋𝜋 + 4 𝑦=5−5 2 + 𝜋𝜋 + 4 𝑦=5 𝜋 + 4 − 2 + 𝜋𝜋 + 4 𝑦= 5 × 2𝜋 + 4 𝑦= 10𝜋 + 4 Hence, for maximum area, Length = 𝒙= 𝟐𝟎𝝅 + 𝟒 m & Breadth = 𝒚= 𝟏𝟎𝝅 + 𝟒 m

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

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 .