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Math Help - Minimize Area from cut wire.....

  1. #1
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    Minimize Area from cut wire.....

    I have been on this problem forever, and cannot solve it. Does anybody have any ideas?

    A wire of length 52 cm is to be cut into two pieces. One of the pieces is to be bent into the form of a circle and the other into the form of a square. How should the wire be cut so that the sum of the enclosed areas is a minimum? What is the perimeter of the square? What is the circumference of the circle?

    Thanks in advance for all the help.
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  2. #2
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    Quote Originally Posted by clayfenderstrat View Post
    I have been on this problem forever, and cannot solve it. Does anybody have any ideas?

    A wire of length 52 cm is to be cut into two pieces. One of the pieces is to be bent into the form of a circle and the other into the form of a square. How should the wire be cut so that the sum of the enclosed areas is a minimum? What is the perimeter of the square? What is the circumference of the circle?

    Thanks in advance for all the help.

    Call the extreme points of wire A, B and call the cut point C ==> the line AC, say of length x, will be shaped bent into a circle and the line CB, of length 52-x, into a square.

    As the circle's perimeter is x=2r\pi, with r=\frac{x}{2\pi}= the circle's radius, the circle's area is \pi\left(\frac{x}{2\pi}\right)^2, and the square's area is (52-x)^2. so the total area of these two shapes is
    \pi\left(\frac{x}{2\pi}\right)^2 + (52-x)^2.
    Now derivate, equal to zero and etc.

    Tonio
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  3. #3
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    Um just one thing, shouldn't the area of the square be ((52-a)/4)^2, as each side of the square is one quarter of the wire left over from the circle.
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  4. #4
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    Quote Originally Posted by chug1 View Post
    Um just one thing, shouldn't the area of the square be ((52-a)/4)^2, as each side of the square is one quarter of the wire left over from the circle.

    Of course you're right! Forgot that "tiny" detail.

    Tonio
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  5. #5
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    When I took the derivative, set it equal to zero, then solved, I ended up with the square having a perimeter of (104/3) and the circle having a circumference of (52/3). However, these answers are incorrect. If someone else could solve this and see if they get the same answers, that would be great. I have worked it about 5 times, and get the same thing every time.

    Thanks
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  6. #6
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    Quote Originally Posted by clayfenderstrat View Post
    When I took the derivative, set it equal to zero, then solved, I ended up with the square having a perimeter of (104/3) and the circle having a circumference of (52/3). However, these answers are incorrect. If someone else could solve this and see if they get the same answers, that would be great. I have worked it about 5 times, and get the same thing every time.

    Thanks
    Show your work and let's see what happened: as \pi appears in the expression for x is odd it doesn't appear above.

    Tonio
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  7. #7
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    Hello, clayfenderstrat!

    Something is wrong . . . Your answers should have \pi in them.


    We have: . A \;=\;\pi\left(\frac{x}{2\pi}\right)^2 + \left(\frac{52-x}{4}\right)^2 \;=\;\pi\left(\frac{x^2}{4\pi^2}\right) + \frac{2704 - 104x + x^2}{16}

    . . . . . . . . A \;=\;\frac{1}{4\pi}\,x^2 + \frac{1}{16}(2704 - 104x + x^2)


    Then: . A' \;=\;\frac{1}{2\pi}\,x + \frac{1}{16}(-104 + 2x) \;=\;0 \quad\Rightarrow\quad \frac{1}{2\pi}\,x - \frac{13}{2} + \frac{1}{8}\,x \;=\;0


    Multiply by 8\pi\!:\;\;4x - 52\pi + \pi x \;=\;0 \quad\Rightarrow\quad \pi x + 4x \;=\;52\pi

    Factor: . (\pi + 4)x \;=\;52\pi \quad\Rightarrow\quad x \;=\;\frac{52\pi}{\pi + 4}


    The circumference of the circle is: . \frac{52\pi}{\pi + 4}\text{ cm.}

    The perimeter of the square is: . \frac{208}{\pi+4}\text{ cm}

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  8. #8
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    Work for Minimization Problem

    y= pi(x/2pi)^2 + ((52-x)/4)^2
    y'= 2pi(x/2pi) - 2 ((52-x)/4)
    x= 2((52/4)-(x/4))
    x=2(13-(2x/4))
    x=26-(x/2)
    x=(52-x)/2
    2x=52-x
    3x=52
    x=(52/3)
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  9. #9
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    Quote Originally Posted by clayfenderstrat View Post
    y= pi(x/2pi)^2 + ((52-x)/4)^2
    y'= 2pi(x/2pi) - 2 ((52-x)/4)

    \color{red}\mbox{This is wrong: you forgot the factor}\,\,\frac{1}{2\pi}\,\,\mbox{from the parentheses in the 1st summand}

    \color{red}\mbox{you also forgot to multiply by}\,\,\frac{1}{4}\,\,\mbox {in the second summand}

    \color{blue}Tonio


    x= 2((52/4)-(x/4))
    x=2(13-(2x/4))
    x=26-(x/2)
    x=(52-x)/2
    2x=52-x
    3x=52
    x=(52/3)
    .
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