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Misc 10 - Sum of perimeter of a circle and square is k - Miscellaneous

Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 3
Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 4
Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 5
Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 6
Misc 10 - Chapter 6 Class 12 Application of Derivatives - Part 7

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.