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Thread: Differentiation

  1. #1
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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 $\displaystyle \frac m4$ long.

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

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



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

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

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

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

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

    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: $\displaystyle R=\frac1{2\pi+8}\cdot d \approx 0.07001239... d$
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