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Misc 11 - A window is a rectangle surmounted by semicircular

Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 4 Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 5 Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 6 Misc 11 - Chapter 6 Class 12 Application of Derivatives - Part 7

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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 Here 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=10/2− 𝑥/2 −1/2×𝜋𝑥/2 𝒚=𝟓−𝒙(𝟏/𝟐+ 𝝅/𝟒) We need to maximize area of window Now, Area of window = Area of rectangle + Area of Semicircle A = Length × Breadth + 1/2 × π𝑟^2 A = 𝑥𝑦+1/2 × 𝜋(𝑥/2)^2 Putting value of y from (1) A = 𝑥(5−𝑥(1/2+ 𝜋/4))+1/2 × (𝜋𝑥^2)/4 A = 5𝑥− 1/2 𝑥^2−(𝜋𝑥^2)/4+(𝜋𝑥^2)/8 A = 𝟓𝒙− 𝟏/𝟐 𝒙^𝟐−(𝝅𝒙^𝟐)/𝟖 Finding 𝒅𝑨/𝒅𝒙 𝑑𝐴/𝑑𝑥=𝑑(5𝑥 − 1/2 𝑥^2 − (𝜋𝑥^2)/8)/𝑑𝑥 𝑑𝐴/𝑑𝑥=5−𝑥−𝜋𝑥/4 Putting 𝒅𝑨/𝒅𝒙=𝟎 0 = 5−𝑥−𝜋𝑥/4 𝑥+𝜋𝑥/4 = 5 (1+𝜋/4)𝑥=5 𝑥=5/((1 + 𝜋/4) ) 𝒙=𝟐𝟎/(𝝅 + 𝟒) Calculating (𝒅^𝟐 𝑨)/(𝒅𝒙^𝟐 ) (𝑑^2 𝐴)/(𝑑𝑥^2 )=𝑑(5 − 𝑥 − 𝜋𝑥/4)/𝑑𝑥 (𝑑^2 𝐴)/(𝑑𝑥^2 )=−1−𝜋/4 <𝟎 So, A’’ <𝟎 at 𝑥=20/(𝜋 + 4) Hence, 𝒙=𝟐𝟎/(𝝅 + 𝟒) maxima Hence, A is maximum when 𝑥=20/(𝜋 + 4) We need to find the dimensions of the window to admit maximum light through the whole opening. 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) ((2 + 𝜋)/4) 𝑦=5− 20/4 ((2 + 𝜋))/(𝜋 + 4) 𝑦=5−5 ((2 + 𝜋))/(𝜋 + 4) 𝑦=5(1−((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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.