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    Member helloying's Avatar
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    problem with finding the minimum area

    A piece of wire of length 8cm is cut into two pieces,one of length x cm,the other of length (8-x)cm.The piece of lenght x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8-x)cm. Show that , as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is \frac{4}{4+\pi}

    i found the area of the square to be (4-\frac{x}{2})cm^2
    and the the radius of the circle is \frac{x}{2\pi}so the area of circle is \frac{x^2}{4\pi}

    i add them up and i differentiate the eqn. i got \frac{x}{2\pi}-0.5

    At min point, \frac{x}{2\pi}-0.5=0
    thus x= \frac{1}{\pi}

    then when i sub it into the radius, i did not get the ans. what is wrong?
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  2. #2
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    Quote Originally Posted by helloying View Post
    A piece of wire of length 8cm is cut into two pieces,one of length x cm,the other of length (8-x)cm.The piece of lenght x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8-x)cm. Show that , as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is \frac{4}{4+\pi}

    i found the area of the square to be (4-\frac{x}{2})cm^2
    and the the radius of the circle is \frac{x}{2\pi}so the area of circle is \frac{x^2}{4\pi}

    i add them up and i differentiate the eqn. i got \frac{x}{2\pi}-0.5

    At min point, \frac{x}{2\pi}-0.5=0
    thus x= \frac{1}{\pi}

    then when i sub it into the radius, i did not get the ans. what is wrong?
    The circumference of the circle is x\,\textrm{cm}, so

    x = 2\pi r

    r = \frac{x}{2\pi}\,\textrm{cm}.


    The area of the circle is \pi r^2 = \pi \left( \frac{x}{2\pi} \right) ^2 = \frac{x^2}{4\pi}\,\textrm{cm}^2.


    The Perimeter of the square is (8 - x)\,\textrm{cm}.

    So 8 - x = 4l

    l = \left(2 - \frac{1}{4}x\right)\,\textrm{cm}.


    Therefore the Area of the Square is \left(2 - \frac{1}{4}x\right)^2 = 4 - x + \frac{1}{16}x^2.


    So the Area of both shapes together is given by...

    A = \frac{1}{4\pi}x^2 + 4 - x + \frac{1}{16}x^2

     = \left(\frac{1}{4\pi} + \frac{1}{16}\right)x^2 - x + 4

     = \frac{4 + \pi}{16\pi}x^2 - x + 4.



    To find the minimum area, we take the derivative, set it equal to 0, and solve for x.

    \frac{dA}{dx} = \frac{4 + \pi}{8\pi}x - 1

    0 = \frac{4 + \pi}{8\pi}x - 1

    1 = \frac{4 + \pi}{8\pi}x

    x = \frac{8\pi}{4 + \pi}.


    We can show we have a minimum by showing that it's second derivative is positive at that point...

    \frac{d^2A}{dx^2} = \frac{4 + \pi}{8\pi} > 0.

    Since the second derivative is positive at that point (and all points), we have a minimum.


    So the minimum occurs when x = \frac{8\pi}{4 + \pi}\,\textrm{cm}.

    Remember that r = \frac{x}{2\pi}\,\textrm{cm}

    r = \frac{\frac{8\pi}{4 + \pi}}{2\pi}

     = \frac{8\pi}{4 + \pi} \times \frac{1}{2\pi}

     = \frac{4}{4 + \pi}\,\textrm{cm}.
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