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MHT CET202613 April 2026Morning ShiftMathematicsApplication of DerivativesActual

A wire 40 metre in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. The lengths of the two pieces so that the combined area of the square and the circle is minimum, are respectively

Options

  1. A160 + 4 , 40 + 4
  2. B+ 4 160 , 40 + 4
  3. C160 + 4 , + 4 40
  4. D30, 10

Correct answer

A. 160 + 4 , 40 + 4

Step-by-step solution

Let the length of the piece for the square be x and for the circle be 40 - x . Perimeter of the square = x side = x 4 Area of the square, A₁ = x^2 16 Circumference of the circle = 40 - x radius r = 40 - x 2 Area of the circle, A₂ = r^2 = (40 - x)^2 4 Total area, A = x^2 16 + (40 - x)^2 4 For minimum area, dA dx = 0 2x 16 - 2(40 - x) 4 = 0 x 8 = 40 - x 2 x = 160 - 4x x( + 4) = 160 x = 160 + 4 Length of the other piece = 40 - x = 40 - 160 + 4 = 40 + 4 Since d^2A dx^2 = 1 8 + 1 2 > 0 , the area is minimum for these le

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