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Math Help - Differentiation

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    Differentiation

    Q:

    A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

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  2. #2
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    Quote Originally Posted by Tangera View Post
    Q:

    A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

    Thank you for helping!
    See here http://www.mathhelpforum.com/math-he...n-problem.html for a similar problem.

    --------------------------
    Let m be the length of the wire that will be transformed into a square.
    Then, d-m will be the length of the wire that will be transformed into a circle.

    The perimeter of the square is m. Therefore, one side of the square is \frac m4 long.

    The area is known to be the square of the side, that is to say :

    \mathcal{A}_s=\left(\frac m4\right)^2=\frac{m^2}{16}



    The perimeter of the circle is d-m. We know that the perimeter of a circle is 2 \pi R.

    --> d-m=2 \pi R \implies R=\frac{d-m}{2 \pi}

    The area of a circle is \mathcal{A}_c=\pi R^2=\pi \left(\frac{d-m}{2 \pi}\right)^2=\frac{(d-m)^2}{4 \pi}

    -------------------
    2 areas are combined :

    \mathcal{A}(m)=\mathcal{A}_s+\mathcal{A}_c

    m is the variable
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  3. #3
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    Quote Originally Posted by Tangera View Post
    Q:

    A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

    Thank you for helping!
    It isn't necessary to help you anymore - Moo did all the interesting parts.

    So only for confirmation: R=\frac1{2\pi+8}\cdot d \approx 0.07001239... d
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