A wire of given length is cut into 2 portions which are divided into the shapes of a circle and square respectively . Show that the sum of the areas of circle and the square will be least when the side of the square is equal to diameter of the circle.
Let the total length of the wire be L. Let a portion x be cut from it and made into a circle. the rest L-x is used to make a square. For the circle, 2R = x R = x/(2) ------------------(1) R being the radius of the circle And for the square, 4a = L-x a = (L-x)/4 -------------------(2)
Sum of areas S = R2 + a2 = x2/4 + (L-x)2/16 For minimum area, dS/dx = 0 which gives x = (L-x)/4 2(x/2) = (L-x)/4 2R = a Hence the sum of the areas of circle and the square will be least when the side of the square is equal to diameter of the circle.
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R = x
) ------------------(1)
