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23 June, 23:43

A piece of wire 40 cm long is to be cut into two pieces. One piece will be bent to form a circle; the other will be bent to form a square.

(a) Find the lengths of the two pieces that cause the sum of the area of the circle and the area of the square to be a minimum.

(b) How could you make the total area of the circle and the square a maximum

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  1. 24 June, 03:08
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    a) 17.6cm and 22.4cm

    b) the total area of the circle and the square can be maximum by differentiating and equating it to zero

    Step-by-step explanation:

    The length of the wire = 40

    Let x be the circumference of the circle.

    x = 2πr

    r = x/2π (r = radius)

    Let the perimeter of the square = (40-x)

    4L = 40-x

    Where L = lenght

    L = (40-x) / 4

    Area of a circle = πr^2

    A = π (x/2π) ^2

    A = π (x^2 / (4π) ^2)

    A = x^2 / 4π

    Area of a square = L^2

    = [ (40-x) / 40]^2

    = (x^2 - 80x + 1600) / 16

    Total area = area of circle + area of square

    = x^2/4π + (x^2 - 80x + 1600) / 16

    = 0.0796x^2 + 0.0625x^2 - 5x + 100

    A = 0.1421x^2 - 5x + 100

    Differentiate A with respect to x

    dA/dx = 0.2842x - 5

    Total area is minimum when dA/dx = 0

    0.2842x - 5 = 0

    0.2842x = 5

    x = 5/0.2842

    x = 17.6cm

    The circumference of the circle is 17.6cm

    40-x = 40 - 17.6 = 22.4cm

    The perimeter of the square is 22.4cm.

    b) the total area of the circle and the square can be maximum by differentiating and equating it to zero
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