Last updated at April 19, 2021 by Teachoo

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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

Miscellaneous

Misc 1
Important
Deleted for CBSE Board 2022 Exams

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important You are here

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc. 19 Deleted for CBSE Board 2022 Exams

Misc 20 Important

Misc 21 Important

Misc 22

Misc. 23 Important

Misc 24 Important

Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.