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Last updated at April 16, 2024 by Teachoo

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 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.