Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12      1. Chapter 6 Class 12 Application of Derivatives
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3. Miscellaneous

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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 be the radius of circle & y be the side of the square Given sum of perimeter of circle & square is k i.e. Circumference of circle + Perimeter of square = k i.e. 2 + 4 = k 2 + 4y = k 4 = k 2 y = 2 4 We have to prove that sum of their area is least when side of square = 2 i.e. we need to minimize the sum of areas Let A be then sum of their area A = Area of circle + Area of square A = 2 + 2 A = 2 + 2 A = 2 + 2 4 2 Differentiating w.r.t = 2 + 2 4 2 = 2 + 2 4 2 = 2 + 1 4 2 2 2 = 2 + 1 16 2 2 . 2 = 2 + 1 16 2 2 . 0 2 = 2 + 1 8 2 2 = 2 2 4 =2 4 2 = 8 + 2 2 4 = 2 4 + 4 Putting =0 2 4 + 4 =0 2 4+ = = 2 4 + = 2 4 + Finding 2 2 = 2 4 + 4 = 2 4 + 4 4 Differentiating w.r.t 2 2 = 2 4 + 4 . 4 2 2 = 2 4 + 4 . 0 2 2 = 2 4 + 4 > 0 2 2 >0 at = 2 4 + = 2 4 + is point of minima Thus, area is least when = 2 4 + From (1) = 2 4 Putting value of = 2 4 + = 2 2 4 + 4 = 4 1 2 2 4 + = 4 1 4 + = 4 4 + + 4 + = 4 4 4 + = 4 + Thus, =2 2 + 4 =2 Hence, proved sum of their area is least when =

Miscellaneous 