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Math Help - A Piece of Wire

  1. #1
    Junior Member Godfather's Avatar
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    A Piece of Wire

    A piece of wire 60 inches long is cut into six sections, two of one length and four of another length. Each of the two sections having the same length is bent into the form of a circle and the two circles are then joined by the four remaining sections to make a frame for a model of a right circular cylinder.
    (a). Find the lengths of the sections which will make the cylinder of maximum volume.
    (b). Justify (a).
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  2. #2
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    Cylinder with maximum volume

    Hello Godfather
    Quote Originally Posted by Godfather View Post
    A piece of wire 60 inches long is cut into six sections, two of one length and four of another length. Each of the two sections having the same length is bent into the form of a circle and the two circles are then joined by the four remaining sections to make a frame for a model of a right circular cylinder.
    (a). Find the lengths of the sections which will make the cylinder of maximum volume.
    (b). Justify (a).
    Suppose the length of one of the two equal lengths is
    x inches. Then:

    • Find, in terms of x, the radius of the circle into which one of these wires can be bent, using circumference of circle = 2\pi r
    • Find the length of one of the four equal lengths, by subtracting 2x from 60 and dividing the result by 4. This gives the height of the cylinder in terms of x.
    • Use the formula for the volume of a cylinder to write down an expression for the volume, V, in terms of x.
    • Use \frac{dV}{dx}=0 to find the value of x which makes the value of V a maximum or minimum.
    • Check that this value gives a maximum.

    Hope you can do it from here.

    Grandad
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    for your first point, if the length of on wire is x then the radius would by r=\frac{x}{2\pi} correct?

    I understand point 3,4 and 5, but not 2.

    Could you please explain 2? Thanks.
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  4. #4
    Eater of Worlds
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    We have 60 inches of wire. 2 pieces are of equal length and 4 of another length.

    Let the 4 pieces be for the height. 60-2x is the total length of those pieces because we allowed the 2x to be for the circular top and bottom.

    There are 4 pieces, so each has length

    \frac{60-2x}{4}=15-\frac{x}{2}

    See now?.
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    thanks a lot,

    So the volume if the the cylinder is v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2}) correct?

    so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?
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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    so i know the derivative is

    \frac{15\pi^3x-\frac{\pi^3x^2}{2}}{2}-\frac{\pi^3x^2}{8}

    but how does one derive that. also, can that be simplified anymore?

    Thanks everyone!
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  7. #7
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    thanks a lot,

    So the volume if the the cylinder is v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2}) correct?

    so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?
    r = \frac{x}{2\pi}

    \frac{dr}{dx} = \frac{1}{2\pi}

    h = 15 - \frac{x}{2}

    \frac{dh}{dx} = -\frac{1}{2}

    V = \pi r^2 h

    \frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot 2r \cdot \frac{dr}{dx} \right]

    now substitute for r, h, and their respective derivatives and optimize.
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  8. #8
    Member OnMyWayToBeAMathProffesor's Avatar
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    how did u get, \frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot 2r \cdot \frac{dr}{dx} \right].

    i thought it would have been \frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot \frac{dr}{dx} \right].
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  9. #9
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    basic implicit derivative ...

    \frac{d}{dx} (r^2) = 2r \frac{dr}{dx}
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  10. #10
    Member OnMyWayToBeAMathProffesor's Avatar
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    i am so sorry, i knew that and shouldn't have missed it.

    so the deriv would be \frac{dV}{dx} = \pi \left[(\frac{x}{2\pi})^2 \cdot \frac{-1}{2} + 15-\frac{x}{2} \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right] = \pi\left[\frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2}\right] right? then i take the deriv of that? which would just be \pi\left[\frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2}\right] ?
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  11. #11
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    i am so sorry, i knew that and shouldn't have missed it.

    so the deriv would be \frac{dV}{dx} = \pi \left[(\frac{x}{2\pi})^2 \cdot \frac{-1}{2} + 15-\frac{x}{2} \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right] = \frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2} right?
    no ... you really need to take more care in dealing with your algebra.

    \frac{dV}{dx} = \pi \left[\left(\frac{x}{2\pi}\right)^2 \cdot \left(-\frac{1}{2}\right) + \left(15-\frac{x}{2}\right) \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right] = \frac{3x(20 - x)}{8\pi}
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  12. #12
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    Differentiating

    Hello OnMyWayToBeAMathProffesor
    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    thanks a lot,

    So the volume if the the cylinder is v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2}) correct?

    so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?
    Sorry to be slow in answering. It seems that you've been going round in circles here! Keep it simple: don't use products, implicit differentiation or chain rule - just multiply out the brackets:

    v = \pi\frac{x^2}{4\pi^2}(15-\frac{x}{2})

    = \frac{15x^2}{4\pi}-\frac{x^3}{8\pi}

    Now differentiate:

    \frac{dv}{dx}=\frac{30x}{4\pi}-\frac{3x^2}{8\pi}

    = \frac{60x-3x^2}{8\pi}

    That was easy, wasn't it?

    Then say \frac{dv}{dx}=0 when 60x-3x^2 = 0

    Factorise and solve for x. There are two answers: one gives a maximum, and one a minimum. OK now?

    Grandad


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  13. #13
    Member OnMyWayToBeAMathProffesor's Avatar
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    Thank you so much Grandad!! and skeeter! so now I'm just going to graph it, find the zeros and check if they are a max or min.

    the 2 zeros are at x=0 and x=20. x=0 is a min and x=20 is a max. thus x=20 is the correct value. Thanks again everybody!

    Happy new years!
    Last edited by OnMyWayToBeAMathProffesor; December 31st 2008 at 09:42 AM.
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