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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

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 2Ο€ (π‘Ήπ’‚π’…π’Šπ’–π’”)+ 4 (π‘Ίπ’Šπ’…π’†) = k 2Ο€π‘₯ + 4y = k 4𝑦 = k – 2Ο€π‘₯ y =(π’Œ βˆ’ πŸπ…π’™)/πŸ’ We need to minimize the Sum of Areas Let A be the Sum of their Area A = Area of circle + Area of square A = Ο€(π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ )^2+(𝑠𝑖𝑑𝑒)^2 A = Ο€π‘₯2 + 𝑦2 A = π𝒙2 + ((π’Œ βˆ’ πŸπ…π’™)/πŸ’)^𝟐 Differentiating w.r.t π‘₯ 𝑑𝐴/𝑑π‘₯=𝑑(γ€–πœ‹π‘₯γ€—^2 + ((π‘˜ βˆ’ 2πœ‹π‘₯)/4)^2 )/𝑑π‘₯ = 𝑑(πœ‹π‘₯^2 )/𝑑π‘₯+𝑑/𝑑π‘₯ ((π‘˜ βˆ’ 2πœ‹π‘₯)/4)^2 = Ο€ 𝑑(π‘₯^2 )/𝑑π‘₯+1/4^2 (𝑑(π‘˜ βˆ’ 2πœ‹π‘₯)^2)/𝑑π‘₯ = Ο€ (2π‘₯)+1/16 2(π‘˜βˆ’2πœ‹π‘₯).𝑑(π‘˜ βˆ’ 2πœ‹π‘₯)/𝑑π‘₯ = 2Ο€ π‘₯ + 1/8 (π‘˜βˆ’2πœ‹π‘₯)(βˆ’2πœ‹) = 2Ο€ π‘₯ – πœ‹(π‘˜ βˆ’ 2πœ‹π‘₯)/4 =2πœ‹π‘₯βˆ’πœ‹/4 (π‘˜βˆ’2πœ‹π‘₯) = (8πœ‹π‘₯ βˆ’ πœ‹π‘˜ + 2πœ‹^2 π‘₯)/4 = (2πœ‹π‘₯(4 + πœ‹) βˆ’ πœ‹π‘˜)/4 Putting 𝒅𝑨/𝒅𝒙=𝟎 (2πœ‹π‘₯(4 + πœ‹) βˆ’ πœ‹π‘˜)/4=0 2Ο€π‘₯(4+πœ‹)=πœ‹π‘˜ π‘₯ = πœ‹π‘˜/2πœ‹(4 + πœ‹) 𝒙 = π’Œ/𝟐(πŸ’ + 𝝅) Finding (𝒅^𝟐 𝑨)/(𝒅𝒙^𝟐 ) 𝑑𝐴/𝑑π‘₯= (2πœ‹π‘₯(4 + πœ‹) βˆ’ πœ‹π‘˜)/4 𝑑𝐴/𝑑π‘₯= 2πœ‹π‘₯(4 + πœ‹)/4 βˆ’ (πœ‹π‘˜ )/4 Differentiating w.r.t π‘₯ (𝑑^2 𝐴)/(𝑑π‘₯^2 ) = 𝑑/𝑑π‘₯ (2πœ‹(4 + πœ‹)/4.π‘₯)βˆ’π‘‘/𝑑π‘₯ (πœ‹π‘˜/4) (𝑑^2 𝐴)/(𝑑π‘₯^2 ) = 2Ο€ ((4 + πœ‹))/4 . 𝑑π‘₯/𝑑π‘₯βˆ’0 (𝑑^2 𝐴)/(𝑑π‘₯^2 ) = 2πœ‹(4 + πœ‹)/4 > 0 Since 𝐀^β€²β€² > 0 for x = π‘˜/2(4 + πœ‹) Thus, A is minimum at x = π‘˜/2(4 + πœ‹) 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 = (π‘˜ βˆ’ 2πœ‹π‘₯)/4 Putting value of π‘₯ = π‘˜/2(4 + πœ‹) 𝑦 = (π‘˜ βˆ’ 2πœ‹(π‘˜/2(4 + πœ‹) ))/4 𝑦 = π‘˜/4 (1βˆ’2πœ‹/2(4 + πœ‹) ) 𝑦 = π‘˜/4 (1βˆ’πœ‹/(4 + πœ‹)) 𝑦 = π‘˜/4 ((4 + πœ‹ + πœ‹)/(4 + πœ‹)) 𝑦 = π‘˜/4 (4/(4 + πœ‹)) π’š = π’Œ/(πŸ’ + 𝝅) Thus, 𝑦=2(π‘˜/2(πœ‹ + 4) ) π’š=πŸπ’™ Hence, Sum of their areas is least when the side of square is double the radius of the circle.

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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.