Calculus

A 30-in. piece of string is cut into two pieces. One piece is used to form a circle and the other to form a square. How shouls the string be cut so that the sum of the areas is a minimum? Round to the nearest tenth, if necessary.

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  1. Let x be the amount used to form the circle, and 30-x is used to form the square. The total enclosed area is
    A = pi x^2 + (30-x)^2
    = (pi+1)x^2 -60 x + 900
    Differentiate that with respect t x and set the derivative equalo to zero.
    2 (pi+1)x -60 = 0
    x = 30/(1 + pi) inches

    Check my work.

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  2. Thanks drwls. I came to the same conclusion after racking my brain over it for some time. I made the problem more difficult than it really was. Thanks again for responding.

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  3. I made two rather bad mistakes. The circular area that can be made piece of length x is
    pi * [x/(2 pi)]^2 and the square area that can be enclosed with length (30-x) is
    [1/4)(30-x)]^2 The total is
    A = x^2/(4 pi) + [7.5 - (x/4)]^2
    = x^2[1/(4 pi) + 1/16] -3.75x +7.5^2

    x [(1/(2 pi) + 1/8] = 3.75
    x[1/pi + 1/4] = 7.5
    x(4/pi + 1) = 30
    x = 30/[(4/pi) + 1]

    Tha maximum area is formed when all of string is used to form a circle. This is a minimum area that can be formed.
    Set dA/dx = 0

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  4. This used to be one of my favourite max/min type of questions when I taught this stuff a long time ago.

    Here is a slightly different approach from the solution done by drwls.

    let the radius of the circle be r,
    then the side of the square is (30-2(pi)r)/4 = (15-(pi)r)/2

    A = [(15-(pi)r)/2]^2 + (pi)r^2

    finding A' and setting this equal to zero gives r = 15/(pi+4)

    then 2(pi)r, the length needed for the circle is 2(pi)(15/(pi+4)) = 30pi/(pi+4) = 13.2 , the same result as drwls

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