# Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,22 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? A wire of length 28m cut into two parts Let be the length of first part Length of 2nd part = 28 Given that one part is converted into a square Let length of m be converted into a square Thus, Perimeter of square = 4 = side of square of = 4 Also, other part is converted into Circle 28 m length converted into a Circle of radius Circumference of circle = 28 m 2 = 28 = 28 2 m We need to find length of pieces so that the combined Area of Circle & Square is minimum Let T be the total area of circle & square T = Area of Circle + Area of Square T = 2 + 2 T = 28 2 2 + 2 16 We have to minimize T Diff T w.r.t = 28 2 4 + 2 16 = 2 28 4 . 28 + 2 16 = 28 2 1 + 8 = 28 2 + 8 Putting =0 28 2 + 8 =0 4 28 + 8 =0 4x 112 + =0 8 4+ =112 = 112 4 + Now finding 2 2 = 28 + 2 + 8 2 2 = 1 2 0+1 + 1 8 2 2 = 1 2 + 1 8 > 0 Hence 2 2 >0 at = 112 4 + Total area is Minimum at = 112 4 + Finding length of other part Length of other part = 28 x = 28 112 4 + = 28 4 + 112 4 + = 28 4 + Length of first part is + & another part is +

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.