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Math Help - 80cm of wire cut into two pieces forming a square and a circle

  1. #1
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    80cm of wire cut into two pieces forming a square and a circle

    Hello everyone.

    Wish i could answer this question on my own,

    The question is from october november 2008 AS level maths paper 1 exam paper

    Question:

    A wire, 80cm long is cut into two pieces. One piece is bent to form a square of side x cm and the other piece is bent to form a circle of radius r cm The total area of the square and circle is A cm^2

    (i) show that A=
    (pi+4)x^2 - 160x + 1600
    -----------------------------
    pi

    I have no idea how to show that...

    then the second question is just as difficult.

    (ii) Given that x and r can vary, find the value of x for which A has a stationary value.

    Would really love some help

    Thanks
    Peder
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  2. #2
    Junior Member slovakiamaths's Avatar
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    India
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    Solution:

    perimeter of square =4x
    perimeter of circle=2 pi r
    then 4x+2 pi r=80
    r=(40-2x)/pi

    Area of square= x^2
    Area of circle= pi r^2

    Total area A= x^2+pir^2
    A= x^2+pi(40-2x/pi)^2
    A= x^2+(1600+4x^2-160x)/pi
    A= (pix^2+1600+4x^2-160x)/pi
    A= [(pi+4)x^2-160x+1600]/pi
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  3. #3
    Super Member

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    Hello, Peder!

    A wire 80cm long is cut into two pieces.
    One piece is bent to form a square of side x cm
    and the other piece is bent to form a circle of radius r cm.
    The total area of the square and circle is A cm³.

    (i) Show that: . A\:=\:\frac{(\pi+4)x^2 - 160x + 1600}{\pi}

    slovakiamath gave an excellent derivation.




    (ii) Given that x and r can vary,
    find the value of x for which A has a stationary value.

    Differentiate A, equate to zero, and solve for x.

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