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Math Help - Question on surface area of cone

  1. #1
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    Question on surface area of cone

    A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 cm^3 of chocolate.

    a) Show that the amount of material used or the curved surface area of the cone is S=\frac{\sqrt{\pi^2 r^6+90000}}{r}

    b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of \pi

    c) The height of the cone in (b) can be expressed as h=\sqrt{ar}, where a is a constant. Find the value of a

    I am able to do (a) and (b), but can't do part (c). Can anyone help??
    Last edited by acc100jt; August 16th 2007 at 07:55 PM.
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  2. #2
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    Quote Originally Posted by acc100jt View Post
    A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 cm^3 of chocolate.

    a) Show that the amount of material used or the curved surface area of the cone is S=\frac{\sqrt{\pi^2 r^6+90000}}{r}

    b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of \pi

    c) The height of the cone in (b) can be expressed as h=\sqrt{ar}, where a is a constant. Find the value of a

    I am able to do (a) and (b), but can't do part (c). Can anyone help??
    This is good for practice.

    Given: right circular cone, V=100 cu.cm, radius r, height h.
    V = pi(h/3)(r^2) = 100
    h = 300/(pi*r^2) --------(1)

    a) The surface area S.
    S = (length of inclined height)(circumference at mid-height of the cone) <--- using Pappus' Centroid Theorem.
    S = (sqrt[r^2 +h^2])(2pi[r/2])
    S = (sqrt[r^2 +90000/(pi^2 *r^4)])(pi*r)
    S = (sqrt[pi^2 *r^6 +90000] /(pi*r^2))(pi*r)
    S = (sqrt[pi^2 *r^6 +90000])/(r) -----------------shown.

    ----------------------------------------------------------------------
    b) Find r when S is minimum.
    So, the r when dS/dr = 0.

    For simplicity, let's simplify S, let's put the denominator r into the radical.
    S = sqrt[(pi^2 *r^4) +(90000/(r^2))]
    dS/dr = (1/2)[4(pi^2)(r^3) -2(90000)/(r^3)] /sqrt[(pi^2 *r^4) +(90000/(r^2))]
    Set that to zero,
    0 = 2(pi^2)(r^3) -90000/(r^3)
    90000 = 2(pi^2)(r^6)
    r^6 = 45000/(pi^2)
    r = [45000/(pi^2)]^(1/6)
    r = 4.072103 cm.

    In terms of pi,
    r = (4.072103)(pi/pi) = 1.29619 pi -----------answer.

    ------------------------------------------------------------
    c) If h in (b) above is sqrt(a*r), find "a".

    h = 300/(pi*r^2) --------(1)
    h = 300/(pi *[(4.072103)^2)]
    h = 5.758825 cm.
    So,
    5.758825 = sqrt[a(4.072103)]
    Square both sides, and then,
    a = 33.164065 / 4.072103
    a = 8.144211 ------------------------answer.
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  3. #3
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    Thanks a lot.
    The answer given in the exercise book for part (c) is 2, i.e. a=2.
    But I couldn't get a=2, now you have confirmed my answer is correct. Thanks!!!
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  4. #4
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    Hello, acc100jt!

    This is how I did part (a) . . .


    A manufacturer intends to use the least amount of material to make
    a hollow right circular cone of radius r cm and vertical height h cm .
    The cone must be able to contain 100 cm³ of chocolate.

    a) Show that the amount of material used or the curved surface area
    of the cone is: . S\:=\:\frac{\sqrt{\pi^2 r^6+90000}}{r}

    The side view of the cone looks like this:
    Code:
                   r
          * - - + - - *
           \    |    /
            \   |   /
             \  |h / R
              \ | /
               \|/
                *
    r is the radius of the cone, h is the height of the cone.
    The slant height of the cone is: . R \:=\:\sqrt{r^2+h^2} .[1]

    Since the volume is 100 cm³: . V \:=\:\frac{1}{3}r^2h \:=\:100\quad\Rightarrow\quad h \:=\:\frac{300}{\pi r^2}

    Substitute into [1]: . R \:=\:\sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2} \:=\:\sqrt{r^2+\frac{90,000}{\pi^2r^4}} :=\:\sqrt{\frac{\pi^2r^6 + 90,000}{\pi^2r^4}}

    . . Hence: . R\:=\:\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2} .[2]


    The cone is formed from a sector of the circle
    . . with radius R and central angle \theta.
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          *                   *
          *         *         *
          *       * θ *       *
                *       * R
           *  *           *  *
            *               *
              *           *
                  * * * 
                    Rθ
    The area of the sector is: . S \:=\:\frac{1}{2}R^2\theta .[3]

    The arclength of the sector (R\theta),
    . . is the circumference of the base of the cone (2\pi r).
    Hence: . R\theta \:=\:2\pi r\quad\Rightarrow \quad\theta \:=\:\frac{2\pi r}{R}

    Substitute into [3]: . S \:=\:\frac{1}{2}R^2\left(\frac{2\pi r}{R}\right)\quad\Rightarrow\quad S \:=\:\pi rR .[4]

    Substitute [2] into [4]: . S \:=\:\pi r\cdot\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2}

    Therefore: . \boxed{S \;=\;\frac{\sqrt{\pi^2r^6 + 90,000}}{r}}

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