Get live Maths 1-on-1 Classs - Class 6 to 12

Miscellaneous

Misc 1 (a)
Deleted for CBSE Board 2023 Exams

Misc 1 (b) Important Deleted for CBSE Board 2023 Exams

Misc 2 Important

Misc 3 Important

Misc 4 Deleted for CBSE Board 2023 Exams

Misc 5 Important Deleted for CBSE Board 2023 Exams

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10 You are here

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc. 19 (MCQ)

Misc 20 (MCQ) Important Deleted for CBSE Board 2023 Exams

Misc 21 (MCQ) Important

Misc 22 (MCQ)

Misc. 23 (MCQ) Important

Misc 24 (MCQ) Important

Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at March 23, 2023 by Teachoo

Misc 10 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.Let 𝒙 be radius of circle & y be side of the square Given Sum of perimeter of circle & square is k Circumference of circle + Perimeter of square = k 2π (𝑹𝒂𝒅𝒊𝒖𝒔)+ 4 (𝑺𝒊𝒅𝒆) = k 2π𝑥 + 4y = k 4𝑦 = k – 2π𝑥 y =(𝒌 − 𝟐𝝅𝒙)/𝟒 We need to minimize the Sum of Areas Let A be the Sum of their Area A = Area of circle + Area of square A = π(𝑟𝑎𝑑𝑖𝑢𝑠)^2+(𝑠𝑖𝑑𝑒)^2 A = π𝑥2 + 𝑦2 A = π𝒙2 + ((𝒌 − 𝟐𝝅𝒙)/𝟒)^𝟐 Differentiating w.r.t 𝑥 𝑑𝐴/𝑑𝑥=𝑑(〖𝜋𝑥〗^2 + ((𝑘 − 2𝜋𝑥)/4)^2 )/𝑑𝑥 = 𝑑(𝜋𝑥^2 )/𝑑𝑥+𝑑/𝑑𝑥 ((𝑘 − 2𝜋𝑥)/4)^2 = π 𝑑(𝑥^2 )/𝑑𝑥+1/4^2 (𝑑(𝑘 − 2𝜋𝑥)^2)/𝑑𝑥 = π (2𝑥)+1/16 2(𝑘−2𝜋𝑥).𝑑(𝑘 − 2𝜋𝑥)/𝑑𝑥 = 2π 𝑥 + 1/8 (𝑘−2𝜋𝑥)(−2𝜋) = 2π 𝑥 – 𝜋(𝑘 − 2𝜋𝑥)/4 =2𝜋𝑥−𝜋/4 (𝑘−2𝜋𝑥) = (8𝜋𝑥 − 𝜋𝑘 + 2𝜋^2 𝑥)/4 = (2𝜋𝑥(4 + 𝜋) − 𝜋𝑘)/4 Putting 𝒅𝑨/𝒅𝒙=𝟎 (2𝜋𝑥(4 + 𝜋) − 𝜋𝑘)/4=0 2π𝑥(4+𝜋)=𝜋𝑘 𝑥 = 𝜋𝑘/2𝜋(4 + 𝜋) 𝒙 = 𝒌/𝟐(𝟒 + 𝝅) Finding (𝒅^𝟐 𝑨)/(𝒅𝒙^𝟐 ) 𝑑𝐴/𝑑𝑥= (2𝜋𝑥(4 + 𝜋) − 𝜋𝑘)/4 𝑑𝐴/𝑑𝑥= 2𝜋𝑥(4 + 𝜋)/4 − (𝜋𝑘 )/4 Differentiating w.r.t 𝑥 (𝑑^2 𝐴)/(𝑑𝑥^2 ) = 𝑑/𝑑𝑥 (2𝜋(4 + 𝜋)/4.𝑥)−𝑑/𝑑𝑥 (𝜋𝑘/4) (𝑑^2 𝐴)/(𝑑𝑥^2 ) = 2π ((4 + 𝜋))/4 . 𝑑𝑥/𝑑𝑥−0 (𝑑^2 𝐴)/(𝑑𝑥^2 ) = 2𝜋(4 + 𝜋)/4 > 0 Since 𝐀^′′ > 0 for x = 𝑘/2(4 + 𝜋) Thus, A is minimum at x = 𝑘/2(4 + 𝜋) We need to prove that Sum of their Areas is least when the side of square is double the radius of the circle So, we need to find value of y "From (1)" 𝑦 = (𝑘 − 2𝜋𝑥)/4 Putting value of 𝑥 = 𝑘/2(4 + 𝜋) 𝑦 = (𝑘 − 2𝜋(𝑘/2(4 + 𝜋) ))/4 𝑦 = 𝑘/4 (1−2𝜋/2(4 + 𝜋) ) 𝑦 = 𝑘/4 (1−𝜋/(4 + 𝜋)) 𝑦 = 𝑘/4 ((4 + 𝜋 + 𝜋)/(4 + 𝜋)) 𝑦 = 𝑘/4 (4/(4 + 𝜋)) 𝒚 = 𝒌/(𝟒 + 𝝅) Thus, 𝑦=2(𝑘/2(𝜋 + 4) ) 𝒚=𝟐𝒙 Hence, Sum of their areas is least when the side of square is double the radius of the circle.