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Last updated at Jan. 7, 2020 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 = 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 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) Now, calculating (𝑑^2 𝐴)/(𝑑𝑥^2 ) at 𝑥=20/(𝜋 + 4) (𝑑^2 𝐴)/(𝑑𝑥^2 )=𝑑(5 − 𝑥 − 𝜋𝑥/4)/𝑑𝑥 (𝑑^2 𝐴)/(𝑑𝑥^2 )=−1−𝜋/4 <0 So, (𝑑^2 𝐴)/(𝑑𝑥^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) ((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
Not in Syllabus - CBSE Exams 2021

Misc 2 Important

Misc 3 Important Not in Syllabus - CBSE Exams 2021

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 Not in Syllabus - CBSE Exams 2021

Misc 20 Important

Misc 21 Important

Misc 22

Misc. 23 Important

Misc 24 Important

Chapter 6 Class 12 Application of Derivatives

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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 9 years. He provides courses for Maths and Science at Teachoo.