Last updated at April 19, 2021 by Teachoo

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 π 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 (πΉππ πππ)+ 4 (πΊππ π) = k 2Οπ₯ + 4y = k 4π¦ = k β 2Οπ₯ y =(π β ππ π)/π 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(πππππ’π )^2+(π πππ)^2 A = Οπ₯2 + π¦2 A = Οπ2 + ((π β ππ π)/π)^π Differentiating w.r.t π₯ ππ΄/ππ₯=π(γππ₯γ^2 + ((π β 2ππ₯)/4)^2 )/ππ₯ = π(ππ₯^2 )/ππ₯+π/ππ₯ ((π β 2ππ₯)/4)^2 = Ο π(π₯^2 )/ππ₯+1/4^2 (π(π β 2ππ₯)^2)/ππ₯ = Ο (2π₯)+1/16 2(πβ2ππ₯).π(π β 2ππ₯)/ππ₯ = 2Ο π₯ + 1/8 (πβ2ππ₯)(β2π) = 2Ο π₯ β π(π β 2ππ₯)/4 =2ππ₯βπ/4 (πβ2ππ₯) = (8ππ₯ β ππ + 2π^2 π₯)/4 = (2ππ₯(4 + π) β ππ)/4 Putting π π¨/π π=π (2ππ₯(4 + π) β ππ)/4=0 2Οπ₯(4+π)=ππ π₯ = ππ/2π(4 + π) π = π/π(π + π ) Finding (π ^π π¨)/(π π^π ) ππ΄/ππ₯= (2ππ₯(4 + π) β ππ)/4 ππ΄/ππ₯= 2ππ₯(4 + π)/4 β (ππ )/4 Differentiating w.r.t π₯ (π^2 π΄)/(ππ₯^2 ) = π/ππ₯ (2π(4 + π)/4.π₯)βπ/ππ₯ (ππ/4) (π^2 π΄)/(ππ₯^2 ) = 2Ο ((4 + π))/4 . ππ₯/ππ₯β0 (π^2 π΄)/(ππ₯^2 ) = 2π(4 + π)/4 > 0 Since π^β²β² > 0 for x = π/2(4 + π) Thus, A is minimum at x = π/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)" π¦ = (π β 2ππ₯)/4 Putting value of π₯ = π/2(4 + π) π¦ = (π β 2π(π/2(4 + π) ))/4 π¦ = π/4 (1β2π/2(4 + π) ) π¦ = π/4 (1βπ/(4 + π)) π¦ = π/4 ((4 + π + π)/(4 + π)) π¦ = π/4 (4/(4 + π)) π = π/(π + π ) Thus, π¦=2(π/2(π + 4) ) π=ππ Hence, Sum of their areas is least when the side of square is double the radius of the circle.

Miscellaneous

Misc 1
Important
Deleted for CBSE Board 2021 Exams only

Misc 2 Important

Misc 3 Important Deleted for CBSE Board 2021 Exams only

Misc 4

Misc 5 Important

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 Important

Misc 17 Important

Misc 18 Important

Misc. 19 Deleted for CBSE Board 2021 Exams only

Misc 20 Important

Misc 21 Important

Misc 22

Misc. 23 Important

Misc 24 Important

About the Author

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.