Now learn Economics at Teachoo for Class 12
Minima/ maxima (statement questions) - Geometry questions
Ex 6.5, 27 (MCQ)
Example 35 Important
Example 41 Important
Example 36
Misc 12 Important
Example 37 Important
Misc 9 Important
Ex 6.5,21
Ex 6.5, 20 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5,22 Important
Misc 10
Misc 11 Important You are here
Ex 6.5,17
Ex 6.5,18 Important
Example 50 Important
Ex 6.5,19 Important
Misc 8 Important
Ex 6.5,23 Important
Misc 15 Important
Misc 17 Important
Example 38 Important
Misc 18 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at April 19, 2021 by Teachoo
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