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Math Help - Optimization: A Wire is to be cut into Two Pieces.

  1. #1
    Junior Member asemh's Avatar
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    Optimization: A Wire is to be cut into Two Pieces.

    Hi

    This question is giving me so much trouble, I mean i do not know how to start it!?

    A wire 70 inches long is to be cut into two pieces. One of the pieces will be bent into the shape of a circle and the other into the shape of an equilateral triangle. Where should the wire be cut in order that the sum of the areas of the circle and the triangle is: a) a minimum? b) a maximum ?


    So far i know that for an Equilateral Triangle all sides is the same so it would be 3x and the remaining will be 70-3x:



    Im really confused as to what Steps I should do to get my answer.

    Thanks
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  2. #2
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    Excellent.

    1) So 70 - 3x is the circumference of your circle. What is the area of that circle.

    2) x is the side of your equilateral triangle. What is the area of that triangle.

    3) Sum?

    4) Min and max of sum?
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  3. #3
    Junior Member asemh's Avatar
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    Am i doing this right so far?



    Then do i derive it and set it to = 0?
    After i did that i got x = 13.24

    Is that correct?
    Please tell me if i am doing anything majorly wrong here, and what i should fix.
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  4. #4
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    Excellent work with finding the area of the circle!!

    Three things come to mind at this point.

    1) "pie" is for dessert. "pi" is for mathematics and Greek Anglicized spellers.

    2) Please seek opportunities to simplify your life. Rather than your last step, notice that \pi is a factor of both Numerator and Denominator and eliminate it, leaving only 4\pi in the Denominator and no \pi at all in the numerator.

    3) Where's the triangle? You must add this to the circle before attempting to minimize the result.

    4) One extra, while I'm thinking about it. Derivatives don't know about endpoints. In a problem such as this, it strikes me as possible that we could lose the triangle or the circle. 3x = 0 or 3x = 70 may be troublesome solutions.
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  5. #5
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    If your circumference is 70 - 3x then

    2\pi r = 70 - 3x

    r = \frac{70 - 3x}{2\pi}.


    Therefore A_{\textrm{circle}} = \pi r^2

     = \pi \left(\frac{70 - 3x}{2\pi}\right)^2

     = \frac{\pi(70 - 3x)^2}{4 \pi ^2}

     = \frac{(70 - 3x)^2}{4\pi}

     = \frac{4900 - 420x + 9x^2}{4\pi}.



    Now you need the area of the triangle.

    Remember that using Heron's Formula

    A = \sqrt{s(s-a)(s-b)(s-c)}, where s = \frac{a + b + c}{2}


    So s = \frac{x + x + x}{2} = \frac{3x}{2}.

    A_{\textrm{triangle}} = \sqrt{s(s - a)(s - b)(s - c)}

     = \sqrt{\frac{3x}{2}\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)}

     = \sqrt{\frac{3x}{2}\left(\frac{x}{2}\right)\left(\f  rac{x}{2}\right)\left(\frac{x}{2}\right)}

     = \sqrt{\frac{3x^4}{8}}

     = \frac{\sqrt{3}x^2}{\sqrt{8}}

     = \frac{\sqrt{3}x^2}{2\sqrt{2}}

     = \frac{\sqrt{6}x^2}{4}.



    So now adding them up:

    A = A_{\textrm{circle}} + A_{\textrm{triangle}}

     = \frac{4900 - 420x + 9x^2}{4\pi} +  \frac{\sqrt{6}x^2}{4}

     = \frac{4900 - 420x + 9x^2}{4\pi} + \frac{\sqrt{6}\pi x^2}{4\pi}

     = \frac{4900 - 420x + (9 + \sqrt{6}\pi)x^2}{4\pi}.


    Now you need to differentiate in order to find the maximum and minimum...


    \frac{dA}{dx} = \frac{-420 + 2(9 + \sqrt{6}\pi)x}{4\pi}

     = \frac{(9 + \sqrt{6}\pi)x - 210}{2\pi}.

    Now set equal to 0 and solve to find the minima and maxima...
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  6. #6
    Junior Member asemh's Avatar
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    "Prove it"

    What exactly did you do in these next 3 steps?

    <br />
= \sqrt{\frac{3x^4}{8}}<br />

    <br />
= \frac{\sqrt{3}x^2}{\sqrt{8}}<br />

    <br />
= \frac{\sqrt{3}x^2}{2\sqrt{2}}<br />

    <br />
= \frac{\sqrt{6}x^2}{4}<br />
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  7. #7
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    Quote Originally Posted by asemh View Post
    "Prove it"

    What exactly did you do in these next 3 steps?

    <br />
= \sqrt{\frac{3x^4}{8}}<br />

    <br />
= \frac{\sqrt{3}x^2}{\sqrt{8}}<br />

    <br />
= \frac{\sqrt{3}x^2}{2\sqrt{2}}<br />

    <br />
= \frac{\sqrt{6}x^2}{4}<br />
    The square root of a fraction is the same as the fraction of the square roots.

    So \sqrt{\frac{3x^4}{8}} = \frac{\sqrt{3x^4}}{\sqrt{8}}

     = \frac{\sqrt{3}\cdot \sqrt{x^4}}{\sqrt{4}\cdot \sqrt{2}}

     = \frac{\sqrt{3}x^2}{2\sqrt{2}}


    And now, rationalising the denominator:

     = \frac{\sqrt{3}x^2 \cdot \sqrt{2}}{2\sqrt{2}\cdot \sqrt{2}}

     = \frac{\sqrt{6}x^2}{4}.
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  8. #8
    Junior Member asemh's Avatar
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    Oh! Yes now I understand
    Thanks
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  9. #9
    Junior Member asemh's Avatar
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    Ok So i set it to Equals 0
    and in the end (Im not sure if i did it right though)
    i got x = 12.6
    So i checked it and know that 12.6 is a minimum.
    How do i find the maximum? =/
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  10. #10
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    Check the end points.

    It also helps if you do a graph of the function.
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