Misc 7 - Chapter 6 Class 12 Application of Derivatives

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Misc 7 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.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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